A body of mass 2 kg rests on a rough inclined plane making anangle 30o...
The body is at rest on the inclined plane, as μ > tan θ force of friction = mg sin 30o = 2 x 9.8 x (1/2) = 9.8 N
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A body of mass 2 kg rests on a rough inclined plane making anangle 30o...
Understanding the Problem
A body of mass 2 kg is resting on a rough inclined plane at an angle of 30 degrees with the horizontal. The coefficient of static friction is given as 0.7. We need to calculate the frictional force acting on the block.
Forces Acting on the Block
- The weight of the block (W) can be calculated as:
- W = mass x gravity = 2 kg x 9.8 m/s² = 19.6 N
- The weight can be resolved into two components:
- Perpendicular to the incline: W⊥ = W * cos(30°)
- Parallel to the incline: W|| = W * sin(30°)
Calculating the Frictional Force
- The maximum static friction (Fs_max) can be calculated using the formula:
- Fs_max = μs * N, where μs is the coefficient of static friction and N is the normal force.
- The normal force (N) on the incline is equal to the perpendicular component of the weight:
- N = W⊥ = W * cos(30°) = 19.6 N * cos(30°) = 19.6 N * (√3/2) ≈ 16.97 N
- Now, substituting the values into the static friction formula:
- Fs_max = 0.7 * N = 0.7 * 16.97 N ≈ 11.88 N
Frictional Force on the Block
- Since the block is at rest, the frictional force will be equal to the force trying to move it down the incline, which is W||.
- W|| = W * sin(30°) = 19.6 N * 0.5 = 9.8 N
Thus, the frictional force acting on the block is equal to the force trying to move it, which is 9.8 N.
Conclusion
The correct answer is option 'A': 9.8 N. The frictional force balances the component of weight parallel to the incline, keeping the block at rest.