Object sliding down a block on a scale

[Lord Anoobis]

Homework Statement

A 200g hamster sits on an 800 g wedge shaped block which in turn rests on a spring scale. An extra fine lubricating oil is sprayed on the top surface of the block, rendering it entirely frictionless, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read as the hamster slides down?

Homework Equations

The Attempt at a Solution

The hamster accelerates down the block with a = gsin$\theta$
The vertical component of this acceleration is:

a_y = gsin²$\theta$

So the downward force exerted by the hamster on the block and thus onto the scale is

F_hamster = ma_y

I added this to the weight of the block and ended up with

Reading = Mg + mgsin²$\theta$
Reading = (0.800)(9.8) + (0.200)(9.8)sin²40^o
Reading = 8.650N

Which is incorrect. I feel that I'm on the right track but I've missed some critical detail. Where did I go wrong?

 

[haruspex]

Try a special case. Suppose theta is zero. What do you get for the force the hamster exerts on the block?

 

[Lord Anoobis]

Try a special case. Suppose theta is zero. What do you get for the force the hamster exerts on the block?

In that case it would simply be $mg$

 

[haruspex]

In that case it would simply be $mg$

Yes, of course it should be, but is that what you get from your mg sin²θ formula?

 

[Lord Anoobis]

Yes, of course it should be, but is that what you get from your mg sin²θ formula?

Indeed not. I can see that mgcos²$\theta$ is correct. What I don't see is how to arrive at that.

 

[Lord Anoobis]

Indeed not. I can see that mgcos²$\theta$ is correct. What I don't see is how to arrive at that.

Wait.

 

[Lord Anoobis]

Yes, of course it should be, but is that what you get from your mg sin²θ formula?

The vertical component of the force the hamster exerts on the block is

F = mgcos²$\theta$

Add this to the weight of the block and there we go.

 

[haruspex]

The vertical component of the force the hamster exerts on the block is

F = mgcos²$\theta$

Add this to the weight of the block and there we go.

Ok. All good now?

 

[Lord Anoobis]

Ok. All good now?

Almost. I would like to look at this a little closer. My initial approach seemed sensible, and I'm not quite sure why it did not work. Logically, that is.

 

[haruspex]

Almost. I would like to look at this a little closer. My initial approach seemed sensible, and I'm not quite sure why it did not work. Logically, that is.

It's because you took the downward acceleration as indicative of the downward force exerted. If you write out the usual $\Sigma F = ma$ equation you'll find you should have negated it.

 

[Lord Anoobis]

It's because you took the downward acceleration as indicative of the downward force exerted. If you write out the usual $\Sigma F = ma$ equation you'll find you should have negated it.

Yes, I see that now. I noticed that subtracting that force from the total weight yielded the correct answer but the reason behind it never clicked. Thanks a whole lot.