Question

Find external force F so that block can move on inclined plane with constant velocity. 

• A. 125 N
• B. 120 N
• C. 145 N
• D. 115 N

Hint:

If \(\mu>\tan \theta\), the block won't slide on its own.
Check if \(\frac{\sqrt{3}}{2}>\tan 30^\circ \implies 0.866>0.577\). Since it is, external help is needed to move it down.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Constant velocity means the net force acting on the block along the incline must be zero.
Forces involved are gravity, friction, and the applied external force.
Step 2: Key Formula or Approach:
1. Force of gravity along incline: \(F_g = mg \sin \theta\).
2. Normal force: \(N = mg \cos \theta\).
3. Kinetic friction: \(f_k = \mu N\).
Step 3: Detailed Explanation:
Given: \(m = 50 \text{ kg}\), \(\theta = 30^\circ\), \(\mu = \frac{\sqrt{3}}{2}\), \(g = 10 \text{ m/s}^2\).
Gravity component down the plane:
\[ mg \sin 30^\circ = 50 \times 10 \times 0.5 = 250 \text{ N} \]
Normal force:
\[ N = mg \cos 30^\circ = 50 \times 10 \times \frac{\sqrt{3}}{2} = 250\sqrt{3} \text{ N} \]
Maximum friction force:
\[ f_k = \mu N = \frac{\sqrt{3}}{2} \times 250\sqrt{3} = \frac{3}{2} \times 250 = 375 \text{ N} \]
Since friction (375 N) is greater than the gravitational component (250 N), the block will not move down by itself.
To move it down with constant velocity, an external force \(F\) must be applied downwards to overcome friction.
\[ F + mg \sin \theta = f_k \]
\[ F + 250 = 375 \]
\[ F = 375 - 250 = 125 \text{ N} \]
Step 4: Final Answer:
The external force required is 125 N.