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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane is
- a)5.4N
- b)10.8N
- c)2.7N
- d)None
Correct answer is option 'B'. Can you explain this answer?
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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. ...
Solution:
Given, weight of cube (W) = 10N
Slope of inclined plane (θ) = 3 in 5
Coefficient of friction (μ) = 0.6
To find: Minimum force required to move the cube up the plane
Normal force (N) acting on the cube is given by:
N = W cosθ
N = 10 cos (tan⁻¹(3/5))
N = 10 × (5/√34)
N = 8.3N
Force of friction (f) acting on the cube is given by:
f = μN
f = 0.6 × 8.3
f = 5N
The force component acting up the inclined plane (F) is given by:
F = W sinθ
F = 10 sin (tan⁻¹(3/5))
F = 10 × (3/√34)
F = 4.9N
Now, to move the cube up the inclined plane, the force acting up the plane should be greater than the force of friction acting down the plane.
i.e., F > f
4.9N > 5N
Therefore, the cube cannot be moved up the plane with this force alone.
To overcome friction, the minimum force required to move the cube up the plane is given by:
Fmin = f + Wsinθ
Fmin = 5 + 4.9
Fmin = 9.9N
Hence, the minimum force required is 10.8N (Option B).
Given, weight of cube (W) = 10N
Slope of inclined plane (θ) = 3 in 5
Coefficient of friction (μ) = 0.6
To find: Minimum force required to move the cube up the plane
Normal force (N) acting on the cube is given by:
N = W cosθ
N = 10 cos (tan⁻¹(3/5))
N = 10 × (5/√34)
N = 8.3N
Force of friction (f) acting on the cube is given by:
f = μN
f = 0.6 × 8.3
f = 5N
The force component acting up the inclined plane (F) is given by:
F = W sinθ
F = 10 sin (tan⁻¹(3/5))
F = 10 × (3/√34)
F = 4.9N
Now, to move the cube up the inclined plane, the force acting up the plane should be greater than the force of friction acting down the plane.
i.e., F > f
4.9N > 5N
Therefore, the cube cannot be moved up the plane with this force alone.
To overcome friction, the minimum force required to move the cube up the plane is given by:
Fmin = f + Wsinθ
Fmin = 5 + 4.9
Fmin = 9.9N
Hence, the minimum force required is 10.8N (Option B).
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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer?
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