What Forces Act on a Box on a Rough Conveyor Belt with Acceleration?

[orochimaru]

Homework Statement

A cubical block of mass \ M and side length \ L 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 \ h above the belt. The conveyer belt then moves to the right with acceleration \ a , 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 \ A and back feet \ B . Hence, determine the value of \ a for which the normal reaction on the front feet is zero.
What happens to the box if this value of \ a is exceeded?

Homework Equations

moment of force = F x r

Newton's laws

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

-Ma

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

would it be \frac{\mu Mg}{2} 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 a or the normal reaction from the equation.

Help would be much appreciated.

 

[Andrew Mason]

You have to first explain why the normal force on the front feet would be zero. (Note: It has nothing to do with the coefficient of static friction).

Think of the belt applying a force to the block at a distance h below the centre of mass. What is the effect of that on the normal force on the front and rear feet?

AM

 

[orochimaru]

thanks for the reply. ok, i think i get what you are saying.

we have resolving vertically: \ Mg = N_A + N_B (1)

and taking moments clockwise about the centre of mass

\ Mah + \frac{N_{A}L}{2} = \frac{N_{B}L}{2} (2)

If I approximate \frac{L}{2} = h because the feet are small
then using equation (1) to substitute for \ N_{A} then \ N_{B} in (2) i get

\ N_{A} = \frac{Mg - 2Ma}{2} and

\ N_{B} = \frac{Mg + 2Ma}{2}

thus when \ N_{A} = 0

\ a = \frac{g}{2}

is this correct?

and if the value of \ a is exceeded the box falls over backwards?

 

[Andrew Mason]

thanks for the reply. ok, i think i get what you are saying.

we have resolving vertically: \ Mg = N_A + N_B (1)

and taking moments clockwise about the centre of mass

\ Mah + \frac{N_{A}L}{2} = \frac{N_{B}L}{2} (2)

If I approximate \frac{L}{2} = h because the feet are small
then using equation (1) to substitute for \ N_{A} then \ N_{B} in (2) i get

\ N_{A} = \frac{Mg - 2Ma}{2} and

\ N_{B} = \frac{Mg + 2Ma}{2}

thus when \ N_{A} = 0

\ a = \frac{g}{2}

is this correct?

and if the value of \ a is exceeded the box falls over backwards?

Looks right. The clockwise torque is Mah. If the normal force on the front feet is 0 then the normal force on the rear is Mg and the counter-clockwise torque is MgL/2. If the total torque is 0, then:

Mg L/2 + Mah = 0 so a = gL/2h

AM