# Homework Help: Object sliding down a block on a scale

1. May 3, 2015

### Lord Anoobis

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

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

ay = gsin2$\theta$

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

Fhamster = may

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)sin240o

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

2. May 3, 2015

### haruspex

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

3. May 3, 2015

### Lord Anoobis

In that case it would simply be $mg$

4. May 3, 2015

### haruspex

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

5. May 3, 2015

### Lord Anoobis

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

6. May 3, 2015

Wait.

7. May 3, 2015

### Lord Anoobis

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.

8. May 3, 2015

### haruspex

Ok. All good now?

9. May 3, 2015

### Lord Anoobis

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.

10. May 3, 2015

### haruspex

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.

11. May 4, 2015

### Lord Anoobis

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.