Question

A crate of mass $m$ rests on the bed of a dump truck. The coefficient of static friction between the crate and the bed of truck is $0.64$. In order to make the crate slide, when the bed is in the position shown, the truck must accelerate to the right. The smallest acceleration $a$ (in terms of gravitational acceleration $g$) for which the crate will begin to slide is

• A. $0.806g$
• B. $g$
• C. $0.0458g$
• D. $0.1667g$

Answer 1

The correct option is C $0.0458g$

This question can quickly be solved by considering pseudo force on m. If the truck is moving with a acceleration in the positive x direction then the pseudo force, ma, will act on m in the negative x direction.

Now, lets calculate the components of the pseudo force:

Force along the slope downward = $macos\theta$

Force normal to the slope upward = $masin\theta$

Frictional force due to contact forces on m

$f=\mu mgcos\theta +\mu masin\theta$

Force along the slope downward (gravitational + pseudo)

$F=mgsin\theta +macos\theta$

When the crate (m) just begin to slide

$F=f$

$⟹mgsin\theta +macos\theta =\mu mgcos\theta +\mu masin\theta$

$⟹a=\frac{\mu cos\theta -sin\theta }{cos\theta +\mu sin\theta }$

Put $\theta ={30}^0$ and $\mu =0.64$

We get, $a=0.04575g$