# Homework Help: Hibbeler 12-210

1. Feb 21, 2014

### c0der

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

Attached

2. Relevant equations

Attached my constraint equations

3. The attempt at a solution

How is sa + sb = d at any stage of motion (circled in the attachment)?

#### Attached Files:

• ###### Hibbeler 12th edition 12-210.png
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2. Feb 21, 2014

### rude man

Pretend just block A moves (B is at rest). Then isn't it obvious A has to move d=3 meters? So sa = d, sb = 0.

Pretend only block B moves. Isn't it obvious block B has to move 3m since A is stationary and d is the distance from the left side of A to the left side of B. So sb = 3 and sa = 0.

So by superposition, if they can both move, sa + sb = d.

3. Feb 21, 2014

### c0der

Thank you, it makes sense that A and B together must travel a distance of 3m so that the right end of B is at the left end of A. However isn't sb+sa misleading as they're measured from the datum? It should be deltaSa + deltaSb = d ?

4. Feb 21, 2014

### rude man

Very good point! But it looks like in the text on the right they decided to make sa and sb both positive by definition, which as you point out is not what the arrows on the diagram on the left define. So, good point but their method still gives the right answer.

In other words, sb should be negative going by the picture arrows but then superposition is sa + (- sb) = sa - sb. Amounts to same thing.

5. Feb 22, 2014

### c0der

I think they mean dsa and dsb but when you integrate from zero to dsa or zero to dsb, they become sa and sb anyway as it's sa-0, sb-0. Either way it makes sense, thanks for your help