# Homework Help: Static eq problem

1. Dec 17, 2006

### lizzyb

Diagram: http://www.webassign.net/pse/p12-11.gif

I'm using a different edition so some of the numbers are different. I'm doing this to review for the exam (it's not for hw).

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

Stephen is pushing his sister Joyce in a wheelbarrow when it is stopped by a brick 8.00 cm high. The handles make an angle of 15.0° from the horizontal. A downward force of 400 N is exerted on the wheel, which has a radius of 20.0 cm.

(a) What force must Stephen apply along the handles to just start the wheel over the brick?

(b) What is the force (magnitude and direction) that the brick exerts on the wheel just as the wheel begins to lift over the brick?

Assume in (a) and (b) that the brick remains fixed and does not slide along the ground.

2. Relevant equations

I guess just the summation of torque?

3. The attempt at a solution

I made a diagram showing the wheel hitting the brick and Stephen's force against the handlebar.

If I place the origin for determining the torque at the axle of the wheel, how can I determine the moment arm of Stephen's force with the given data?

2. Dec 17, 2006

### OlderDan

If the wheel is going to climb the brick, the wheel is going to rotate around the point of contact. Consider the forces acting on the wheel.

In my opinion the problem is a bit ambiguous about the force on the wheel. The 400N may or may not include a force Stephen is exerting downward. He has to be lifting the handles as well as pushing forward. My guess is you should only worry about the horizontal force he applies, and assume the 400N downward is constant in the problem. The only upward force to overcome the 400N will be from the brick. The problem with this is that the 15 degrees would not matter.

Last edited: Dec 17, 2006
3. Dec 17, 2006

### lizzyb

Forces acting on the wheel:
• Force of brick against wheel (P)
• Weight of the wheelbarrow (W)
• Stephen's force pushing the wheelbarrow (F)

So if it's going to rotate about the point of contact, I should take the origin at that place.

4. Dec 17, 2006

### OlderDan

Yes.. the geometry of the wheel must be considered to find the relationships among these forces. I just saw the picture at the other link. I need to think a bit more about what the problem is expecting for an answer, but you should be able to find the horizontal component of Stephen's force in any case.

Last edited: Dec 17, 2006
5. Dec 17, 2006

### lizzyb

For the weight of the wheelbarrow, we have the moment arm d as: $$d_{w} = \sqrt{2 r h - h^2}$$. But surely Stephen's force is applicable as torque?

Let phi be the angle from the horizontal the force R (from the brick) makes upon the wheel.

F_x = F cos(15) - R cos(phi) = m a
F_y = R sin(phi) - mg - F sin(15) = 0 (??)

The solution is: (a) 859N, (b) 1040 N left and upward at 36.9 degrees.

The question (a) states "What force must Stephen apply along the handles to just start the wheel over the brick?"

So if we take the origin as the center of wheel, we don't need to use Stephen's force in determining the torque:

tau = ????

Last edited: Dec 17, 2006
6. Dec 17, 2006

### OlderDan

OK- I think that they intend for you to neglect the upward force Stephen would be applying before encountering the brick. Part of that force is "along the handles" but you cannot possibly know that force from the information given. I assume they do want you to include the additional downward force (actually a reduction in the upward force) Stephen applies in the torque calculation. So make the assumption that Stephen is pushing both forward and downward at the 15º angle. The net downward force at the wheel axle will be the original 400N plus the added downward component of Stephen's push at 15º, and the horizontal force will be the horizontal component of Stephen's push. Nothing is moving until the wheel starts to lift off the ground, so all forces are balanced.

The torque about the axle is not what determines when the wheel starts to lift. For that you need the torque about the point of contact with the brick. At the axle there is a force that has a horizontal component, caused by stephens pushing and a vertical component caused partly by Stephen's pushing and partly by the original 400N. With gentle to no pushing, the torque about the contact point from the vertical component will be greater than the torque from the horizontal component. As Stephen pushes harder, the torque from the horizontal component will increase faster than that from the vertical component. Eventually, they will become equal and then the wheel will start to lift.

7. Dec 17, 2006

### OlderDan

Is this your solution or the book's answer? It is correct. The angle phi is the angle the resultant force makes with the horizontal when it is pointing at the axle, which can be determined from the geometry. This is the point at which the torque from the downward force equals the torque from the horizontal force in magnitude.

tau_total = 0 = (400N + Fsin15)(.16m) - Fcos15(.12m)

8. Dec 17, 2006

### lizzyb

it is the book's answer; i'm still trying to take what you wrote in. thanks for your help.

9. Dec 17, 2006

### lizzyb

where is the .12 m from in:

tau_total = 0 = (400N + Fsin15)(.16m) - Fcos15(.12m)

I understand the F cos 15 is the horizontal part of Stephen's push, but where did the moment arm come from?

OOPS / Edit
Sorry - yes, the .12 m = 20 cm - 8 cm.

10. Dec 17, 2006

### lizzyb

got it! thanks!!! :-)