# Determine the Acceleration of the of the Initally Stationary Block

1. Apr 20, 2013

### Northbysouth

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

Determine the acceleration a of the initially stationary 15-kg body when the 33-N force P is applied as shown. The small wheels at B are ideal, and the feet at A are small.

I have attached an image of the problem

2. Relevant equations

3. The attempt at a solution

I started with summing the forces in the x and y direction:

ƩFx: max = P - Bx - F

ƩFy: may = By -mg + A

Then I took the moment about B to get:

ƩMB =0 = -hP +.5lmg - lA

A = [0.5mg - hP]/l

A = 52.895 N

From here I am then able to determine the value of By by using my equation for the sum of the forces in the y-direction

By = 94.255 N

I know that Fmax = usA

Hence Fmax = 23.2738 N

But how do I then determine Bx and F? I believe that I need to ensure the block is actually moving, but Bx remains unknown so I cannot determine whether F is greater than its maximum value.

I had tried taking the moment about another point, the top right corner of the box, which gave me an equation containing Bx and F, which I had then solved for F so I could plug it into my ƩFx equation, but I feel as though Bx would vary depending on whether the block was in motion or not, because the value of F would change.

Can someone clarify this for me?

Thank you

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2. Apr 20, 2013

### Simon Bridge

By "l" you mean "L" in your diagram?

... it is usually better practice to leave the number crunching to the end.
Leave $AL=\frac{1}{2}mg - hP$.

For #B_x## ... you know that point B is an ideal wheel. What is the horizontal resistance of an ideal wheel?
F is friction - in order to just get the contraption moving you must have just overcome which kind of friction?

3. Apr 20, 2013

### Northbysouth

You are correct in the assumption that "l" is equal to "L" in the diagram.

I recognize that there is no Bx force, somehow I confused the wheel with a pin, not quite sure how I managed to do that...

Also, I recognize that F = ukA

Hence, shouldn't I get:

ax = [P-ukA]/L

which gives me 20.73 m/s2

but it says its wrong and I don't understand where my mistake is. Is there something wrong with how I calculated A by summing the moments about B?

4. Apr 20, 2013

### Simon Bridge

You can check by treating as a free-body and finding out the sum of forces through the center of mass.

5. Apr 20, 2013

### Northbysouth

So, I looked at my calculations again; this time I took the moment about the center of the block. In doing this I was able to find the value of A and hence I correctly calculated the acceleration to be a = 0.9392 m/s^2.

Why did I get an incorrect value for A when I took the moment about B? What makes this scenario different to any earlier material I've studied?

Also, could you expand on what you mean by finding the sum of the forces through the center mass, please?

6. Apr 21, 2013

### Simon Bridge

In a free body, any force acting on the body can be split into components through the center of mass and perpendicular to it. The perpendicular component rotates the body. The other one translates it. You were interested in the translational acceleration.

You should be able to figure out which effect you missed out by comparing the free-body equations with the previous ones. See how many of your original terms can be extracted. What's left will tell you what happened.