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Crane problem (reaction forces)

by JamesL
Tags: crane, forces, reaction
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JamesL
#1
Mar30-04, 09:28 PM
P: 33
Im having a bit of trouble with this problem:

A crane of mass 5288.51 kg supports a load of 789.67 kg. The crane are is 7m long and the angle it makes wiht the horizontal is 47.627 degrees. The distance between the front and rear wheels is 4 m.

Assume that the center of mass of the crane, is at the center of the crane and that the crane arm is of negligible mass. The reaction force at the rear wheels is Nr and at the front is Nf. The sum of these is Nt (total reaction force at the ground).

Find Nt and Nr.

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There is a drawing to go along with this.... it makes it easier to visualize... the crane is basically a rectangle sitting on the ground. The rear wheel is just represented by the back of the rectangle and the front wheel by the front. The cranes arm is sticking out of the front at the bottom.

If anybody could point me in the right direction i would greatly appreciate it!
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paul11273
#2
Mar30-04, 10:37 PM
P: 156
You should have two unknows, the reaction at each wheel. So this will require you to have two equations.
Let one equation be the summation of forces in the y-axis (down) and the second equation be the summation of the moments about any point. I would pick a conveniant point, probably the front wheel. That would cause Nf to drop out.
Solve these two equations for Nr.
JamesL
#3
Mar30-04, 11:28 PM
P: 33
Thanks for the response.

This is what ive got so far:

M = mass of crane
m = mass of load

Nr + Nf - Mg - mg = 0 for the forces in the y direction

... im not really sure how to set up the equation around the front wheel (although that does seem like the point to do it at).... any ideas?

paul11273
#4
Mar31-04, 08:16 AM
P: 156
Crane problem (reaction forces)

Come to think of it, I do not think you need the equation for the forces in the y direction. When you set up the summation of the moment about the front wheel, then Nf will drop from the equation. That will leave only Nr to solve for.

However, unless there is another way to solve this, there is missing information in your problem. You will need to know the distance of the CG of the crane from the front wheel. In the description of the problem, I have a picture in my head that the crane body is modeled as a rectangle (or square) with the CG in the center, and the crane arm extends from the front of it. Your moment equation will be something like:

d1= distance of CG from front wheel
d2= distance from front wheel to rear wheel
L=length of the crane arm
let positive rotation be counter clockwise (typical Cartesian convention of a positive angle)
Mf: (d*Mg)+(d2*Nr)-(mg*L*cos(47.627))=0

Post what the distances are and I will try the calculations too, to see if our numbers agree.

Edit - P.S. I kept the same mass assignments you used, M=mass of crane and m=mass of hanging weight.


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