Practice Test - MCQs test series for Term 2 Exams

Solve the following question:

Asked by nkrbasak | 29th Dec, 2011, 09:37: PM

Expert Answer:

we will chose the x axis to be along the wedge since motion is possible in this direction only. Let us call the force required to just push it up the place $F_1$ and the force required to just prevent it from sliding down as $F_2$ . We need to draw the free body diagrams in both cases:

In the first case, friction acts to prevent $F_1$ from driving the mass up the plane. In the second case, friction helps $F_2$ to balance out gravitational force. In both cases

$mg\cos\theta=N$

Balancing the forces along the x axis for both cases (remember that there is no acceleration hence net force is zero)

$F_1 = mg\sin\theta + \mu N = mg(\sin\theta+\mu\cos\theta)$

$F_2 = mg\sin\theta - \mu N = mg(\sin\theta-\mu\cos\theta)$

Dividing the equations, we get

$\frac{F_1}{F2}=\frac{(\sin\theta+\mu\cos\theta)}{(\sin\theta-\mu\cos\theta)}=\frac{(\tan\theta+\mu)}{(\tan\theta-\mu)}=3$

Solve to get $\mu =0.5$ and N = 5. Keep in mind that this N has nothing to do with the normal force. This is simply a quantity defined in the question as 10 times $\mu$

Answered by | 29th Dec, 2011, 10:29: PM

Ask Doubt