Question

A block placed on a horizontal surface is being pushed by a force F making an angle $\theta$ with the vertical. If the friction coefficient is $\mu$, how much force is needed to get the block just started.

• A. F=$\frac{\mu mg}{sin\theta -\mu cos\theta }$
• B. F=$\frac{\mu mg}{\mu cos\theta -sin\theta }$
• C. F=$\mu$mg(sin$\theta$ - $\mu$cos$\theta$)
• D. F=$\mu$mg($\mu$cos$\theta$ - sin$\theta$)

Answer 1

The correct option is A

F=$\frac{\mu mg}{sin\theta -\mu cos\theta }$

The situation is shown in figure.In the limiting equillibrium the frictional force f will be equal to $\mu N$.

For horizontal equillibrium

F sin$\theta$ = $\mu N$

For vertical equillibrium

Fcos$\theta$ + mg =N

Eliminating N from these equations,

Fsin$\theta$=$\mu$Fcos$\theta$ +$\mu$mg

$F=\frac{\mu mg}{sin\theta -\mu cos\theta }$