# Mech Help - Moment & Friction

1. Homework Statement

http://img237.imageshack.us/img237/7580/mechhelpui7.jpg [Broken]

2. The attempt at a solution

(i) Moment of F = 70 x 800 = 56000
Moment of weight = 500 x 270 = 135000

Therefore the trolley remains at rest. Is this right?

(ii) Can you help me with this question please?

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# Hi
# for (i) Yes you are right. The thing will not start to rotate until the moment due to force exceeds the moment of weight.
# (ii) For calculating the moment one looks at the perpendicular distance between the force vector and the pivot point and they are the same for figure I and II. but the pivot points are different. A in I and B in II. Hope this helps.

Hi, thank you for confirming (i) is correct.

Im still a little confused about (ii). It gives me a coefficient of friction value of 0.5 for the wheel. I know:

F = Coefficient of friction X Resultant
F = 0.5 x 500
F = 250N = Friction force acting in the opposite direction of the force pushing the trolley

Is this corrent so far? I dont really know where to go from here, could anyone help me out?

Doc Al
Mentor
2. The attempt at a solution

(i) Moment of F = 70 x 800 = 56000
Moment of weight = 500 x 270 = 135000

Therefore the trolley remains at rest. Is this right?
What about the forces at B?

Am i not accounting for the weight at B by finding the moment of the weight force? If im doing something wrong can you please set me straight, its the first question on this past exam paper and its driving me nuts.

Doc Al
Mentor
Am i not accounting for the weight at B by finding the moment of the weight force?
No. There is a reaction force of the ground on the object at point B. Figure out that normal force. Then you can compare the applied force F to the static friction force.

Since the weight is central, does that mean the weight on point A and point B will be equal? If this is the case, that means there is a reaction force of 250N at point B.

Is that correct?

Doc Al
Mentor
No. You can't just forget about the applied force F. Consider moments of all forces to find the normal force.

Force F as in the friction force on the wheel or the force pushing the trolley?

I really appreciate your help Doc Al, but im having a hard time understanding the process i have to go through to get the rigth answer. Can you possibly go into a little more detail so i am able to understand?

Doc Al
Mentor
Force F as in the friction force on the wheel or the force pushing the trolley?
Force F as in the applied force on the object. (Look at the diagram!)

The wheels are frictionless; friction will only act at point B.

Draw yourself a diagram showing all forces acting on the object. Don't neglect the forces acting at point B.

Hint: Consider moments about point A.