(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A cubical block of mass [tex]\ M [/tex] and side length [tex]\ L [/tex] with small feet attached to the 4 corners of its base is placed on a rough conveyer belt initially at rest. The centre of mass of the box is midway between the front and rear feet and is a height [tex]\ h [/tex] above the belt. The conveyer belt then moves to the right with acceleration [tex]\ a [/tex] , the box remaining at rest relative to the belt.

Indicate the real forces on the box. Also indicate the ficticious force which when introduced allows you to consider the box to be in equilibrium in its own frame of reference.

Determine the normal reaction on the front feet [tex]\ A [/tex] and back feet [tex]\ B [/tex] . Hence, determine the value of [tex]\ a [/tex] for which the normal reaction on the front feet is zero.

What happens to the box if this value of [tex]\ a [/tex] is exceeded?

2. Relevant equations

moment of force = F x r

Newton's laws

3. The attempt at a solution

I know this needs a diagram really. Is there any way I can put one here?

Anyway, the real forces are

1/ normal reaction of front feet

2/ normal reaction of back feet

3/ Mg acting down from centre of mass

4/ Friction between front and back feet and the ground.

The ficticious force is

[tex] -Ma [/tex]

I'm a bit confused which way the friction acts and its magnitude

would it be [tex]\frac{\mu Mg}{2} [/tex] the thing is they don't give the co-efficient of friction in the question

I think the solution is to take moments about a point and since it is in equilibrium the resultant moment should be 0 but wherever I take moments I seem to eliminate [tex] a [/tex] or the normal reaction from the equation.

Help would be much appreciated.

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# Homework Help: Mechanics: box on conveyer belt

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