Friction - 2
1
A block of mass m is sliding down a rough inclined plane at an angle of 60° with an acceleration of g/2. What is the coefficient of kinetic friction μ_k between the block and the inclined plane?
Friction <-- Kinetic Friction on Inclined Plane
A
μ_k = 1/2
B
μ_k = √3 - 1
C
μ_k = √3/2
D
μ_k = 1
✅ Show Answer
✔ Correct Answer:
B
(μ_k = √3 - 1)
💡 Explanation
Given:
Inclination angle, θ = 60°
Acceleration of block, a = g/2
The block is sliding downward, so kinetic friction acts upward along the incline.
Step 1: Identify the forces along the incline.
Component of gravitational force acting down the incline:
mg sinθ
Kinetic friction acts upward:
f_k = μ_kN
The normal reaction is:
N = mg cosθ
Therefore:
f_k = μ_kmg cosθ
Step 2: Apply Newton's second law along the incline.
Taking downward along the incline as positive:
mg sinθ - f_k = ma
Substitute f_k = μ_kmg cosθ:
mg sinθ - μ_kmg cosθ = ma
Step 3: Substitute the given acceleration.
a = g/2
Therefore:
mg sin60° - μ_kmg cos60° = mg/2
Cancel mg from both sides:
sin60° - μ_k cos60° = 1/2
Using:
sin60° = √3/2
cos60° = 1/2
We get:
√3/2 - μ_k/2 = 1/2
Multiply by 2:
√3 - μ_k = 1
Therefore:
μ_k = √3 - 1
Numerically:
μ_k ≈ 0.732
Hence, the coefficient of kinetic friction is √3 - 1.
Inclination angle, θ = 60°
Acceleration of block, a = g/2
The block is sliding downward, so kinetic friction acts upward along the incline.
Step 1: Identify the forces along the incline.
Component of gravitational force acting down the incline:
mg sinθ
Kinetic friction acts upward:
f_k = μ_kN
The normal reaction is:
N = mg cosθ
Therefore:
f_k = μ_kmg cosθ
Step 2: Apply Newton's second law along the incline.
Taking downward along the incline as positive:
mg sinθ - f_k = ma
Substitute f_k = μ_kmg cosθ:
mg sinθ - μ_kmg cosθ = ma
Step 3: Substitute the given acceleration.
a = g/2
Therefore:
mg sin60° - μ_kmg cos60° = mg/2
Cancel mg from both sides:
sin60° - μ_k cos60° = 1/2
Using:
sin60° = √3/2
cos60° = 1/2
We get:
√3/2 - μ_k/2 = 1/2
Multiply by 2:
√3 - μ_k = 1
Therefore:
μ_k = √3 - 1
Numerically:
μ_k ≈ 0.732
Hence, the coefficient of kinetic friction is √3 - 1.
🎯 Conclusion
How to think about this problem:
1. Since the block is sliding downward, kinetic friction acts upward along the incline.
2. The downward component of gravity is mg sinθ.
3. The normal reaction is mg cosθ, so kinetic friction is μ_kmg cosθ.
4. Apply Newton's second law along the incline:
mg sinθ - μ_kmg cosθ = ma.
5. Substitute θ = 60° and a = g/2.
6. Simplifying gives:
μ_k = √3 - 1 ≈ 0.732.
Key formula for a block sliding down a rough incline:
a = g(sinθ - μ_k cosθ)
Here:
a = g/2
θ = 60°
Therefore:
μ_k = √3 - 1.
1. Since the block is sliding downward, kinetic friction acts upward along the incline.
2. The downward component of gravity is mg sinθ.
3. The normal reaction is mg cosθ, so kinetic friction is μ_kmg cosθ.
4. Apply Newton's second law along the incline:
mg sinθ - μ_kmg cosθ = ma.
5. Substitute θ = 60° and a = g/2.
6. Simplifying gives:
μ_k = √3 - 1 ≈ 0.732.
Key formula for a block sliding down a rough incline:
a = g(sinθ - μ_k cosθ)
Here:
a = g/2
θ = 60°
Therefore:
μ_k = √3 - 1.
2
A 50 kg body is moving on a horizontal surface on Earth in the +x direction. The coefficient of kinetic friction between the body and surface is μ_k = 0.3. What is the kinetic frictional force acting on the body? (Take g = 10 m/s²)
Friction <-- Kinetic Friction on Horizontal Surface
A
f_k = 50 N in the -x direction
B
f_k = 100 N in the -x direction
C
f_k = 150 N in the -x direction
D
f_k = 300 N in the -x direction
✅ Show Answer
✔ Correct Answer:
C
(f_k = 150 N in the -x direction)
💡 Explanation
Given:
Mass of body, m = 50 kg
Coefficient of kinetic friction, μ_k = 0.3
Acceleration due to gravity, g = 10 m/s²
The body is moving in the +x direction.
Step 1: Determine the normal reaction.
Since the surface is horizontal and there are no other vertical forces:
N = mg
N = 50 × 10 = 500 N
Step 2: Calculate the kinetic friction.
The magnitude of kinetic friction is:
f_k = μ_kN
f_k = 0.3 × 500
f_k = 150 N
Step 3: Determine the direction of friction.
Friction always acts opposite to the direction of relative motion.
The body is moving in the +x direction, so friction acts in the -x direction.
Therefore:
f⃗_k = -150 î N
The magnitude of kinetic friction is 150 N, directed toward the -x direction.
Mass of body, m = 50 kg
Coefficient of kinetic friction, μ_k = 0.3
Acceleration due to gravity, g = 10 m/s²
The body is moving in the +x direction.
Step 1: Determine the normal reaction.
Since the surface is horizontal and there are no other vertical forces:
N = mg
N = 50 × 10 = 500 N
Step 2: Calculate the kinetic friction.
The magnitude of kinetic friction is:
f_k = μ_kN
f_k = 0.3 × 500
f_k = 150 N
Step 3: Determine the direction of friction.
Friction always acts opposite to the direction of relative motion.
The body is moving in the +x direction, so friction acts in the -x direction.
Therefore:
f⃗_k = -150 î N
The magnitude of kinetic friction is 150 N, directed toward the -x direction.
🎯 Conclusion
How to think about this problem:
1. For a body moving on a horizontal surface:
N = mg.
2. Calculate the normal force:
N = 50 × 10 = 500 N.
3. Calculate kinetic friction:
f_k = μ_kN = 0.3 × 500 = 150 N.
4. Friction opposes motion. Since the body moves in the +x direction, friction acts in the -x direction.
Therefore:
f_k = 150 N toward -x.
In vector form:
f⃗_k = -150 î N.
1. For a body moving on a horizontal surface:
N = mg.
2. Calculate the normal force:
N = 50 × 10 = 500 N.
3. Calculate kinetic friction:
f_k = μ_kN = 0.3 × 500 = 150 N.
4. Friction opposes motion. Since the body moves in the +x direction, friction acts in the -x direction.
Therefore:
f_k = 150 N toward -x.
In vector form:
f⃗_k = -150 î N.
3
A body of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 s due to friction between the body and the floor. What is the coefficient of friction? (Take acceleration due to gravity g = 10 m/s²)
Friction <-- Kinetic Friction and Retardation
A
μ = 0.2
B
μ = 0.3
C
μ = 0.4
D
μ = 0.5
✅ Show Answer
✔ Correct Answer:
C
(μ = 0.4)
💡 Explanation
Given:
Mass of body, m = 10 kg
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Time taken to stop, t = 5 s
Acceleration due to gravity, g = 10 m/s²
Step 1: Calculate the acceleration of the body.
Using the kinematic equation:
v = u + at
Substitute the given values:
0 = 20 + a(5)
Therefore:
5a = -20
a = -4 m/s²
The negative sign indicates that the acceleration is opposite to the direction of motion.
The magnitude of deceleration is therefore 4 m/s².
Step 2: Relate friction to acceleration.
Since the body is moving on a horizontal floor:
N = mg
Kinetic friction is:
f_k = μ_kN
f_k = μ_kmg
Using Newton's second law, the frictional force produces the deceleration:
f_k = m|a|
Therefore:
μ_kmg = m|a|
Cancel m from both sides:
μ_kg = |a|
Substitute the values:
μ_k(10) = 4
Therefore:
μ_k = 4/10 = 0.4
Hence, the coefficient of kinetic friction between the body and the floor is 0.4.
Mass of body, m = 10 kg
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Time taken to stop, t = 5 s
Acceleration due to gravity, g = 10 m/s²
Step 1: Calculate the acceleration of the body.
Using the kinematic equation:
v = u + at
Substitute the given values:
0 = 20 + a(5)
Therefore:
5a = -20
a = -4 m/s²
The negative sign indicates that the acceleration is opposite to the direction of motion.
The magnitude of deceleration is therefore 4 m/s².
Step 2: Relate friction to acceleration.
Since the body is moving on a horizontal floor:
N = mg
Kinetic friction is:
f_k = μ_kN
f_k = μ_kmg
Using Newton's second law, the frictional force produces the deceleration:
f_k = m|a|
Therefore:
μ_kmg = m|a|
Cancel m from both sides:
μ_kg = |a|
Substitute the values:
μ_k(10) = 4
Therefore:
μ_k = 4/10 = 0.4
Hence, the coefficient of kinetic friction between the body and the floor is 0.4.
🎯 Conclusion
How to think about this problem:
1. The body stops because friction produces a constant deceleration.
2. First find the acceleration using:
v = u + at.
This gives a = -4 m/s².
3. On a horizontal surface, friction is:
f_k = μ_kmg.
4. Newton's second law gives:
f_k = m|a|.
5. Equating the two:
μ_kmg = m|a|.
6. The mass cancels:
μ_k = |a|/g.
7. Therefore:
μ_k = 4/10 = 0.4.
Key idea:
For a body sliding on a horizontal surface under kinetic friction alone, the magnitude of deceleration is μ_kg.
1. The body stops because friction produces a constant deceleration.
2. First find the acceleration using:
v = u + at.
This gives a = -4 m/s².
3. On a horizontal surface, friction is:
f_k = μ_kmg.
4. Newton's second law gives:
f_k = m|a|.
5. Equating the two:
μ_kmg = m|a|.
6. The mass cancels:
μ_k = |a|/g.
7. Therefore:
μ_k = 4/10 = 0.4.
Key idea:
For a body sliding on a horizontal surface under kinetic friction alone, the magnitude of deceleration is μ_kg.
4
A block of mass 5 kg is placed at rest on a table of rough surface. A force of 30 N is applied parallel to the surface of the table. The block slides through a distance of 50 m in an interval of time 10 s. What is the coefficient of kinetic friction? (Take g = 10 m/s²)
Friction <-- Kinetic Friction and Kinematics
A
μ_k = 0.2
B
μ_k = 0.4
C
μ_k = 0.5
D
μ_k = 0.6
✅ Show Answer
✔ Correct Answer:
C
(μ_k = 0.5)
💡 Explanation
Given:
Mass of block, m = 5 kg
Applied force, F = 30 N
Initial velocity, u = 0 m/s
Distance travelled, s = 50 m
Time taken, t = 10 s
Acceleration due to gravity, g = 10 m/s²
Step 1: Calculate the acceleration of the block.
Using the kinematic equation:
s = ut + (1/2)at²
Since the block starts from rest, u = 0:
50 = 0 + (1/2)a(10)²
50 = 50a
Therefore:
a = 1 m/s²
Step 2: Calculate the net force on the block.
Using Newton's second law:
F_net = ma
F_net = 5 × 1 = 5 N
The applied force is 30 N, while the net force is only 5 N. Therefore, kinetic friction opposes the applied force.
F_net = F - f_k
5 = 30 - f_k
Therefore:
f_k = 25 N
Step 3: Calculate the normal reaction.
The applied force is parallel to the horizontal surface, so it has no vertical component.
Therefore:
N = mg
N = 5 × 10 = 50 N
Step 4: Calculate the coefficient of kinetic friction.
Kinetic friction is:
f_k = μ_kN
Therefore:
25 = μ_k × 50
μ_k = 25/50
μ_k = 0.5
Hence, the coefficient of kinetic friction is 0.5.
Mass of block, m = 5 kg
Applied force, F = 30 N
Initial velocity, u = 0 m/s
Distance travelled, s = 50 m
Time taken, t = 10 s
Acceleration due to gravity, g = 10 m/s²
Step 1: Calculate the acceleration of the block.
Using the kinematic equation:
s = ut + (1/2)at²
Since the block starts from rest, u = 0:
50 = 0 + (1/2)a(10)²
50 = 50a
Therefore:
a = 1 m/s²
Step 2: Calculate the net force on the block.
Using Newton's second law:
F_net = ma
F_net = 5 × 1 = 5 N
The applied force is 30 N, while the net force is only 5 N. Therefore, kinetic friction opposes the applied force.
F_net = F - f_k
5 = 30 - f_k
Therefore:
f_k = 25 N
Step 3: Calculate the normal reaction.
The applied force is parallel to the horizontal surface, so it has no vertical component.
Therefore:
N = mg
N = 5 × 10 = 50 N
Step 4: Calculate the coefficient of kinetic friction.
Kinetic friction is:
f_k = μ_kN
Therefore:
25 = μ_k × 50
μ_k = 25/50
μ_k = 0.5
Hence, the coefficient of kinetic friction is 0.5.
🎯 Conclusion
How to think about this problem:
1. First find acceleration from the given distance and time:
s = ut + (1/2)at².
This gives a = 1 m/s².
2. Find the net force using Newton's second law:
F_net = ma = 5 N.
3. The applied force is 30 N, so the remaining 25 N must be kinetic friction:
f_k = 30 - 5 = 25 N.
4. Since the force is parallel to the horizontal surface:
N = mg = 50 N.
5. Use:
f_k = μ_kN.
Therefore:
μ_k = 25/50 = 0.5.
Key idea:
When both force and displacement information are given, first use kinematics to find acceleration, then Newton's second law to find friction, and finally f_k = μ_kN to find the coefficient.
1. First find acceleration from the given distance and time:
s = ut + (1/2)at².
This gives a = 1 m/s².
2. Find the net force using Newton's second law:
F_net = ma = 5 N.
3. The applied force is 30 N, so the remaining 25 N must be kinetic friction:
f_k = 30 - 5 = 25 N.
4. Since the force is parallel to the horizontal surface:
N = mg = 50 N.
5. Use:
f_k = μ_kN.
Therefore:
μ_k = 25/50 = 0.5.
Key idea:
When both force and displacement information are given, first use kinematics to find acceleration, then Newton's second law to find friction, and finally f_k = μ_kN to find the coefficient.
5
A block of mass 10 kg is lying on a horizontal surface and is pulled by a force F acting at an angle of 30° with the horizontal. If the coefficient of static friction is μ_s = 0.25, what is the value of F for which the block will just start moving? (Take g = 10 m/s²)
Friction <-- Limiting Friction with Inclined Pull
A
33.3 N
B
25.2 N
C
20 N
D
35.7 N
✅ Show Answer
✔ Correct Answer:
B
(25.2 N)
💡 Explanation
Given:
Mass of block, m = 10 kg
Angle of applied force, θ = 30° above the horizontal
Coefficient of static friction, μ_s = 0.25
Acceleration due to gravity, g = 10 m/s²
The block is just about to move, so the static friction has reached its maximum value:
f_s = f_s,max = μ_sN
Step 1: Resolve the applied force into horizontal and vertical components.
Horizontal component:
F_x = F cos30°
Vertical component:
F_y = F sin30°
The vertical component acts upward and therefore reduces the normal reaction.
Step 2: Calculate the normal reaction.
Since there is no vertical acceleration:
N + F sin30° - mg = 0
Therefore:
N = mg - F sin30°
N = 100 - F/2
Step 3: Apply the condition for impending motion.
At the instant the block is about to move, the horizontal component of the applied force equals maximum static friction:
F cos30° = μ_sN
Substitute the values:
F(√3/2) = 0.25(100 - F/2)
Therefore:
0.866F = 25 - 0.125F
0.991F = 25
F ≈ 25.2 N
Therefore, the required pulling force is approximately 25.2 N.
Mass of block, m = 10 kg
Angle of applied force, θ = 30° above the horizontal
Coefficient of static friction, μ_s = 0.25
Acceleration due to gravity, g = 10 m/s²
The block is just about to move, so the static friction has reached its maximum value:
f_s = f_s,max = μ_sN
Step 1: Resolve the applied force into horizontal and vertical components.
Horizontal component:
F_x = F cos30°
Vertical component:
F_y = F sin30°
The vertical component acts upward and therefore reduces the normal reaction.
Step 2: Calculate the normal reaction.
Since there is no vertical acceleration:
N + F sin30° - mg = 0
Therefore:
N = mg - F sin30°
N = 100 - F/2
Step 3: Apply the condition for impending motion.
At the instant the block is about to move, the horizontal component of the applied force equals maximum static friction:
F cos30° = μ_sN
Substitute the values:
F(√3/2) = 0.25(100 - F/2)
Therefore:
0.866F = 25 - 0.125F
0.991F = 25
F ≈ 25.2 N
Therefore, the required pulling force is approximately 25.2 N.
🎯 Conclusion
How to think about this problem:
1. The block is just about to move, so friction is maximum static friction.
2. Resolve the applied force into horizontal and vertical components.
Horizontal component = F cos30°.
Upward component = F sin30°.
3. The upward component reduces the normal force:
N = mg - F sin30°.
4. Maximum static friction is:
f_s,max = μ_sN.
5. At impending motion, horizontal force equals maximum static friction:
F cos30° = μ_sN.
6. Solving gives:
F ≈ 25.2 N.
Key idea:
When a block is pulled upward at an angle, the vertical component of the pulling force reduces the normal force. Therefore, the limiting friction is smaller than it would be for a purely horizontal force.
1. The block is just about to move, so friction is maximum static friction.
2. Resolve the applied force into horizontal and vertical components.
Horizontal component = F cos30°.
Upward component = F sin30°.
3. The upward component reduces the normal force:
N = mg - F sin30°.
4. Maximum static friction is:
f_s,max = μ_sN.
5. At impending motion, horizontal force equals maximum static friction:
F cos30° = μ_sN.
6. Solving gives:
F ≈ 25.2 N.
Key idea:
When a block is pulled upward at an angle, the vertical component of the pulling force reduces the normal force. Therefore, the limiting friction is smaller than it would be for a purely horizontal force.
6
The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. What is the minimum magnitude of horizontal force that should be applied on the block to keep it adhered to the wall? (Take g = 10 m/s²)
Friction <-- Vertical Wall
A
5 N
B
10 N
C
25 N
D
50 N
✅ Show Answer
✔ Correct Answer:
C
(25 N)
💡 Explanation
Given:
Mass of block, m = 0.5 kg
Coefficient of static friction, μ_s = 0.2
Acceleration due to gravity, g = 10 m/s²
The block is pressed horizontally against a vertical wall.
Step 1: Determine the normal reaction.
The applied horizontal force presses the block against the wall.
Since there is no horizontal acceleration:
N = F
Step 2: Determine the maximum static friction.
The maximum static friction is:
f_s,max = μ_sN
f_s,max = μ_sF
f_s,max = 0.2F
Step 3: Balance the vertical forces.
The weight of the block acts downward:
mg = 0.5 × 10 = 5 N
For the block to be just able to remain adhered to the wall, the maximum friction must balance its weight:
f_s,max = mg
Therefore:
0.2F = 5
F = 5/0.2
F = 25 N
Therefore, the minimum horizontal force required is 25 N.
Mass of block, m = 0.5 kg
Coefficient of static friction, μ_s = 0.2
Acceleration due to gravity, g = 10 m/s²
The block is pressed horizontally against a vertical wall.
Step 1: Determine the normal reaction.
The applied horizontal force presses the block against the wall.
Since there is no horizontal acceleration:
N = F
Step 2: Determine the maximum static friction.
The maximum static friction is:
f_s,max = μ_sN
f_s,max = μ_sF
f_s,max = 0.2F
Step 3: Balance the vertical forces.
The weight of the block acts downward:
mg = 0.5 × 10 = 5 N
For the block to be just able to remain adhered to the wall, the maximum friction must balance its weight:
f_s,max = mg
Therefore:
0.2F = 5
F = 5/0.2
F = 25 N
Therefore, the minimum horizontal force required is 25 N.
🎯 Conclusion
How to think about this problem:
1. The block tends to slide downward because of gravity.
2. Static friction acts upward to prevent the downward motion.
3. The horizontal force creates the normal reaction from the wall:
N = F.
4. Maximum static friction is:
f_s,max = μ_sN = μ_sF.
5. At the minimum force needed to keep the block from falling:
μ_sF = mg.
6. Substitute the values:
0.2F = 0.5 × 10.
Therefore:
F = 25 N.
Key idea:
For a block pressed against a vertical wall, the horizontal applied force determines the normal force, and the resulting friction supports the block against gravity.
1. The block tends to slide downward because of gravity.
2. Static friction acts upward to prevent the downward motion.
3. The horizontal force creates the normal reaction from the wall:
N = F.
4. Maximum static friction is:
f_s,max = μ_sN = μ_sF.
5. At the minimum force needed to keep the block from falling:
μ_sF = mg.
6. Substitute the values:
0.2F = 0.5 × 10.
Therefore:
F = 25 N.
Key idea:
For a block pressed against a vertical wall, the horizontal applied force determines the normal force, and the resulting friction supports the block against gravity.
7
A body of mass 1 kg rests on a horizontal floor with coefficient of static friction μ_s = 1/√3. It is desired to make the body move by applying the minimum possible force F. What is the value of F? (Take g = 10 m/s² and round off to the nearest integer.)
Friction <-- Minimum Force on Rough Horizontal Surface
A
3 N
B
5 N
C
7 N
D
10 N
✅ Show Answer
✔ Correct Answer:
B
(5 N)
💡 Explanation
Given:
Mass of body, m = 1 kg
Coefficient of static friction, μ_s = 1/√3
Acceleration due to gravity, g = 10 m/s²
To obtain the minimum possible force, the force should be applied at an angle θ above the horizontal.
Step 1: Resolve the applied force into components.
Horizontal component = F cosθ
Vertical component = F sinθ
The vertical component acts upward and reduces the normal reaction.
Therefore:
N = mg - F sinθ
Step 2: Apply the condition for impending motion.
At the instant the body is about to move:
F cosθ = μ_sN
Substitute N = mg - F sinθ:
F cosθ = μ_s(mg - F sinθ)
Rearranging:
F(cosθ + μ_s sinθ) = μ_smg
Therefore:
F = μ_smg / (cosθ + μ_s sinθ)
Step 3: Find the angle that gives minimum F.
The denominator is maximum when:
tanθ = μ_s
Since:
μ_s = 1/√3
we get:
tanθ = 1/√3
θ = 30°
Step 4: Calculate the minimum force.
F_min = (μ_smg) / (cos30° + μ_s sin30°)
Substitute the values:
F_min = [(1/√3)(1)(10)] / [(√3/2) + (1/√3)(1/2)]
The denominator is:
√3/2 + 1/(2√3) = 2/√3
Therefore:
F_min = (10/√3) / (2/√3)
F_min = 5 N
Hence, the minimum force required to just start the motion is 5 N.
Mass of body, m = 1 kg
Coefficient of static friction, μ_s = 1/√3
Acceleration due to gravity, g = 10 m/s²
To obtain the minimum possible force, the force should be applied at an angle θ above the horizontal.
Step 1: Resolve the applied force into components.
Horizontal component = F cosθ
Vertical component = F sinθ
The vertical component acts upward and reduces the normal reaction.
Therefore:
N = mg - F sinθ
Step 2: Apply the condition for impending motion.
At the instant the body is about to move:
F cosθ = μ_sN
Substitute N = mg - F sinθ:
F cosθ = μ_s(mg - F sinθ)
Rearranging:
F(cosθ + μ_s sinθ) = μ_smg
Therefore:
F = μ_smg / (cosθ + μ_s sinθ)
Step 3: Find the angle that gives minimum F.
The denominator is maximum when:
tanθ = μ_s
Since:
μ_s = 1/√3
we get:
tanθ = 1/√3
θ = 30°
Step 4: Calculate the minimum force.
F_min = (μ_smg) / (cos30° + μ_s sin30°)
Substitute the values:
F_min = [(1/√3)(1)(10)] / [(√3/2) + (1/√3)(1/2)]
The denominator is:
√3/2 + 1/(2√3) = 2/√3
Therefore:
F_min = (10/√3) / (2/√3)
F_min = 5 N
Hence, the minimum force required to just start the motion is 5 N.
🎯 Conclusion
How to think about this problem:
1. To minimize the required force, do not apply the force horizontally. Apply it at an angle above the horizontal.
2. The upward component F sinθ reduces the normal reaction:
N = mg - F sinθ.
3. Reduced normal force means reduced maximum static friction.
4. At impending motion:
F cosθ = μ_sN.
5. The required force is minimized when:
tanθ = μ_s.
6. Here:
μ_s = 1/√3, so θ = 30°.
7. Substituting θ = 30° gives:
F_min = 5 N.
Key idea:
For the minimum pulling force on a rough horizontal surface, the optimum angle satisfies tanθ = μ_s.
1. To minimize the required force, do not apply the force horizontally. Apply it at an angle above the horizontal.
2. The upward component F sinθ reduces the normal reaction:
N = mg - F sinθ.
3. Reduced normal force means reduced maximum static friction.
4. At impending motion:
F cosθ = μ_sN.
5. The required force is minimized when:
tanθ = μ_s.
6. Here:
μ_s = 1/√3, so θ = 30°.
7. Substituting θ = 30° gives:
F_min = 5 N.
Key idea:
For the minimum pulling force on a rough horizontal surface, the optimum angle satisfies tanθ = μ_s.
8
A block of mass 10 kg is kept on a rough inclined plane at 45°. A force of 3 N is applied to the block upward along the incline. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P such that the block does not move downward? (Take g = 10 m/s²)
Friction <-- Inclined Plane with Additional Force
A
18 N
B
25 N
C
32 N
D
23 N
✅ Show Answer
✔ Correct Answer:
B
(25 N)
💡 Explanation
Given:
Mass of block, m = 10 kg
Inclination angle, θ = 45°
Force acting upward along the incline = 3 N
Additional force P acts upward along the incline
Coefficient of static friction, μ_s = 0.6
Acceleration due to gravity, g = 10 m/s²
We need the minimum value of P such that the block does not move downward.
Step 1: Calculate the component of weight along the incline.
The component of gravity acting down the incline is:
mg sin45°
= 10 × 10 × (1/√2)
= 50√2 N
Step 2: Calculate the normal reaction.
Since all applied forces are parallel to the incline, they have no perpendicular component.
Therefore:
N = mg cos45°
= 100 × (1/√2)
= 50√2 N
Step 3: Calculate maximum static friction.
f_s,max = μ_sN
= 0.6 × 50√2
= 30√2 N
Step 4: Determine the direction of friction.
We want the minimum P such that the block does not move downward.
At the limiting condition, the block is just about to move downward.
Therefore, static friction acts upward along the incline.
Step 5: Apply force balance along the incline.
Upward forces:
P + 3 + 30√2
Downward force:
50√2
At the limiting condition:
P + 3 + 30√2 = 50√2
Therefore:
P = 20√2 - 3
P ≈ 28.28 - 3
P ≈ 25.28 N
The nearest integer value is:
P = 25 N.
Mass of block, m = 10 kg
Inclination angle, θ = 45°
Force acting upward along the incline = 3 N
Additional force P acts upward along the incline
Coefficient of static friction, μ_s = 0.6
Acceleration due to gravity, g = 10 m/s²
We need the minimum value of P such that the block does not move downward.
Step 1: Calculate the component of weight along the incline.
The component of gravity acting down the incline is:
mg sin45°
= 10 × 10 × (1/√2)
= 50√2 N
Step 2: Calculate the normal reaction.
Since all applied forces are parallel to the incline, they have no perpendicular component.
Therefore:
N = mg cos45°
= 100 × (1/√2)
= 50√2 N
Step 3: Calculate maximum static friction.
f_s,max = μ_sN
= 0.6 × 50√2
= 30√2 N
Step 4: Determine the direction of friction.
We want the minimum P such that the block does not move downward.
At the limiting condition, the block is just about to move downward.
Therefore, static friction acts upward along the incline.
Step 5: Apply force balance along the incline.
Upward forces:
P + 3 + 30√2
Downward force:
50√2
At the limiting condition:
P + 3 + 30√2 = 50√2
Therefore:
P = 20√2 - 3
P ≈ 28.28 - 3
P ≈ 25.28 N
The nearest integer value is:
P = 25 N.
🎯 Conclusion
How to think about this problem:
1. The block tends to slide downward because of the component of gravity along the incline.
2. The 3 N force and P both act upward along the incline and help prevent downward motion.
3. For the minimum P, the block is just about to move downward, so static friction acts upward at its maximum value.
4. Calculate the weight component along the incline:
mg sin45° = 50√2 N.
5. Calculate the normal reaction:
N = mg cos45° = 50√2 N.
6. Maximum static friction:
f_s,max = μ_sN = 30√2 N.
7. Balance forces along the incline:
P + 3 + 30√2 = 50√2.
8. Therefore:
P = 20√2 - 3 ≈ 25.28 N.
Hence, the nearest integer answer is:
P = 25 N.
Key idea:
When finding the minimum additional force needed to prevent downward motion, friction acts upward and reaches its maximum value at the limiting condition.
1. The block tends to slide downward because of the component of gravity along the incline.
2. The 3 N force and P both act upward along the incline and help prevent downward motion.
3. For the minimum P, the block is just about to move downward, so static friction acts upward at its maximum value.
4. Calculate the weight component along the incline:
mg sin45° = 50√2 N.
5. Calculate the normal reaction:
N = mg cos45° = 50√2 N.
6. Maximum static friction:
f_s,max = μ_sN = 30√2 N.
7. Balance forces along the incline:
P + 3 + 30√2 = 50√2.
8. Therefore:
P = 20√2 - 3 ≈ 25.28 N.
Hence, the nearest integer answer is:
P = 25 N.
Key idea:
When finding the minimum additional force needed to prevent downward motion, friction acts upward and reaches its maximum value at the limiting condition.
9
A body of mass 2 kg slides down a rough inclined plane of angle 30° with an acceleration of 3 m/s². What external force parallel to the incline is required to take the same body up the plane with the same acceleration? (Take g = 10 m/s²)
Friction <-- Kinetic Friction on Inclined Plane
A
10 N
B
14 N
C
4 N
D
20 N
✅ Show Answer
✔ Correct Answer:
D
(20 N)
💡 Explanation
Given:
Mass of body, m = 2 kg
Inclination angle, θ = 30°
Acceleration while sliding downward, a = 3 m/s²
Acceleration due to gravity, g = 10 m/s²
Step 1: Determine the kinetic friction acting on the body.
While the body slides downward, friction acts upward along the incline.
Apply Newton's second law along the incline:
mg sinθ - f_k = ma
Substitute the given values:
2 × 10 × sin30° - f_k = 2 × 3
10 - f_k = 6
Therefore:
f_k = 4 N
Step 2: Determine the force required to move the body upward.
Now the body is to move upward with the same acceleration (3 m/s²).
Both gravity and kinetic friction act downward along the incline.
Apply Newton's second law along the incline:
F - mg sinθ - f_k = ma
Substitute the values:
F - 10 - 4 = 2 × 3
F - 14 = 6
Therefore:
F = 20 N
Hence, the required external force is 20 N.
Mass of body, m = 2 kg
Inclination angle, θ = 30°
Acceleration while sliding downward, a = 3 m/s²
Acceleration due to gravity, g = 10 m/s²
Step 1: Determine the kinetic friction acting on the body.
While the body slides downward, friction acts upward along the incline.
Apply Newton's second law along the incline:
mg sinθ - f_k = ma
Substitute the given values:
2 × 10 × sin30° - f_k = 2 × 3
10 - f_k = 6
Therefore:
f_k = 4 N
Step 2: Determine the force required to move the body upward.
Now the body is to move upward with the same acceleration (3 m/s²).
Both gravity and kinetic friction act downward along the incline.
Apply Newton's second law along the incline:
F - mg sinθ - f_k = ma
Substitute the values:
F - 10 - 4 = 2 × 3
F - 14 = 6
Therefore:
F = 20 N
Hence, the required external force is 20 N.
🎯 Conclusion
How to think about this problem:
1. First analyze the downward motion.
Gravity pulls the body downward, while kinetic friction opposes the motion.
2. Apply Newton's second law:
mg sinθ - f_k = ma.
This gives:
f_k = 4 N.
3. Now consider the upward motion.
Since the body moves upward, both gravity and friction act downward.
4. Apply Newton's second law again:
F - mg sinθ - f_k = ma.
5. Substitute the known values:
F - 10 - 4 = 6.
6. Therefore:
F = 20 N.
Key idea:
Whenever the direction of motion reverses, kinetic friction also reverses its direction. Always redraw the free-body diagram before writing the force equation.
1. First analyze the downward motion.
Gravity pulls the body downward, while kinetic friction opposes the motion.
2. Apply Newton's second law:
mg sinθ - f_k = ma.
This gives:
f_k = 4 N.
3. Now consider the upward motion.
Since the body moves upward, both gravity and friction act downward.
4. Apply Newton's second law again:
F - mg sinθ - f_k = ma.
5. Substitute the known values:
F - 10 - 4 = 6.
6. Therefore:
F = 20 N.
Key idea:
Whenever the direction of motion reverses, kinetic friction also reverses its direction. Always redraw the free-body diagram before writing the force equation.
10
A block of mass 5 kg is placed on a rough inclined surface as shown. The inclination of the plane is 30° and the coefficient of static friction is μ_s = 0.1. If F_1 is the force required to just move the block up the inclined plane and F_2 is the force required to just prevent the block from sliding down, what is the value of |F_1 - F_2|? (Take g = 10 m/s²)
Friction <-- Limiting Static Friction on Inclined Plane
A
5√3 N
B
5√3/2 N
C
50√3 N
D
10 N
✅ Show Answer
✔ Correct Answer:
A
(5√3 N)
💡 Explanation
Given:
Mass of block, m = 5 kg
Inclination angle, θ = 30°
Coefficient of static friction, μ_s = 0.1
Acceleration due to gravity, g = 10 m/s²
Step 1: Calculate the normal reaction.
Since the applied forces F_1 and F_2 act parallel to the incline:
N = mg cosθ
N = 5 × 10 × cos30°
N = 50 × (√3/2)
N = 25√3 N
Step 2: Calculate the maximum static friction.
f_s,max = μ_sN
f_s,max = 0.1 × 25√3
f_s,max = 5√3/2 N
Step 3: Find F_1.
F_1 is the force required to just move the block upward.
Therefore, friction acts downward along the incline.
For limiting upward motion:
F_1 = mg sinθ + f_s,max
F_1 = 5 × 10 × sin30° + 5√3/2
F_1 = 25 + 5√3/2 N
Step 4: Find F_2.
F_2 is the force required to just prevent the block from sliding downward.
Therefore, friction acts upward along the incline.
For limiting downward motion:
F_2 + f_s,max = mg sinθ
Therefore:
F_2 = mg sinθ - f_s,max
F_2 = 25 - 5√3/2 N
Step 5: Calculate |F_1 - F_2|.
|F_1 - F_2| = |(25 + 5√3/2) - (25 - 5√3/2)|
|F_1 - F_2| = |5√3|
|F_1 - F_2| = 5√3 N
Therefore, the required answer is 5√3 N.
Mass of block, m = 5 kg
Inclination angle, θ = 30°
Coefficient of static friction, μ_s = 0.1
Acceleration due to gravity, g = 10 m/s²
Step 1: Calculate the normal reaction.
Since the applied forces F_1 and F_2 act parallel to the incline:
N = mg cosθ
N = 5 × 10 × cos30°
N = 50 × (√3/2)
N = 25√3 N
Step 2: Calculate the maximum static friction.
f_s,max = μ_sN
f_s,max = 0.1 × 25√3
f_s,max = 5√3/2 N
Step 3: Find F_1.
F_1 is the force required to just move the block upward.
Therefore, friction acts downward along the incline.
For limiting upward motion:
F_1 = mg sinθ + f_s,max
F_1 = 5 × 10 × sin30° + 5√3/2
F_1 = 25 + 5√3/2 N
Step 4: Find F_2.
F_2 is the force required to just prevent the block from sliding downward.
Therefore, friction acts upward along the incline.
For limiting downward motion:
F_2 + f_s,max = mg sinθ
Therefore:
F_2 = mg sinθ - f_s,max
F_2 = 25 - 5√3/2 N
Step 5: Calculate |F_1 - F_2|.
|F_1 - F_2| = |(25 + 5√3/2) - (25 - 5√3/2)|
|F_1 - F_2| = |5√3|
|F_1 - F_2| = 5√3 N
Therefore, the required answer is 5√3 N.
🎯 Conclusion
How to think about this problem:
1. The component of weight down the incline is mg sinθ.
2. The normal reaction is N = mg cosθ.
3. Maximum static friction is μ_sN.
4. For impending upward motion, friction acts downward:
F_1 = mg sinθ + μ_smg cosθ.
5. For impending downward motion, friction acts upward:
F_2 = mg sinθ - μ_smg cosθ.
6. Subtracting the two equations:
F_1 - F_2 = 2μ_smg cosθ.
7. Substituting the values:
|F_1 - F_2| = 2(0.1)(5)(10)cos30°
= 5√3 N.
Key idea:
The difference between the forces needed for impending upward and downward motion is twice the maximum static friction.
1. The component of weight down the incline is mg sinθ.
2. The normal reaction is N = mg cosθ.
3. Maximum static friction is μ_sN.
4. For impending upward motion, friction acts downward:
F_1 = mg sinθ + μ_smg cosθ.
5. For impending downward motion, friction acts upward:
F_2 = mg sinθ - μ_smg cosθ.
6. Subtracting the two equations:
F_1 - F_2 = 2μ_smg cosθ.
7. Substituting the values:
|F_1 - F_2| = 2(0.1)(5)(10)cos30°
= 5√3 N.
Key idea:
The difference between the forces needed for impending upward and downward motion is twice the maximum static friction.