Trecherous incline

Classical Mechanics Level 4

Two friends decide to shove boxes up a rough plank inclined at an angle of $\theta$ . The plank is slightly smoother at the bottom and a bit rougher at the top, such that the coefficient of kinetic friction increases linearly with the distance $s$ along the plank: $\mu_k=ks$ . One of the friends shoves a box up the plank so that it leaves the bottom of the plank at a speed of $V_{0}$ . Assuming that the coefficients of kinetic and static friction are equal $\mu_k=\mu_s$ , when the box first comes to rest, it will remain at rest if $(V_{0})^A \geq \frac{B \times g \times sin^C(\theta)}{k \times cos^D(\theta)}$ For some constant positive integers $A,B,C,D$ ; what is the value of $A+B+C+D$ ?

Details and Assumptions

$g$ is gravitational acceleration

The answer is 8.

2 solutions

Satvik Pandey
Nov 10, 2014

Block will reach equilibrium if

$mgsin\theta =\mu mgcos\theta$

on putting the value of $\mu$ we will get

$s=\frac { tan\theta }{ k }$

Now net force acting on the block in direction parallel to plane is $mgsin\theta +\mu mgcos\theta$

So $m\frac { dv }{ dt } =mgsin\theta +\mu mgcos\theta$

or $v\frac { dv }{ dx } =gsin\theta +ksgcos\theta$

or $\int _{ { V }_{ 0 } }^{ 0 }{ vdv } =\int _{ 0 }^{ \frac { tan\theta }{ k } }{ (gsin\theta +ksgcos\theta )ds }$

On solving we will get

${ V }_{ 0 }^{ 2 }\ge \frac { 3g{ sin }^{ 2 }\theta }{ kcos\theta }$

So $A+B+C+D=8$

The problem actually depends on the length of the box :\ I tried working it out with an arbitrary length L, but it turned out dependent on L

But, assuming the length is negligible, I get the same solution as yours :)

Nathanael Case - 6 years, 5 months ago

OH! yes it depends on the length of the box. I didn't think about that. But as no information is given about the length. So we can assume it to be negligible. :D

satvik pandey - 6 years, 5 months ago

The problem would be a lot more fun if it took the length in to account... I want to know if my solution is correct!

It kind of bothers me because, if the length is negligible, then the box should be tiny enough to carry! Hahaha :)

Nathanael Case - 6 years, 5 months ago

Vivek Bhagat
Oct 15, 2014

Eqn becomes simple if you apply work energy theorem and for the equilibrium position do the free body diagram, substitute value of distance traveled obtained from fbd in the work energy theorem and you will get the answer. Unlike many problems here frictional force depends on distance traveled, so you will have to intrgrate the frictional force into distance eqn to obtained work done by frictional force

Trecherous incline

The answer is 8.

2 solutions

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