Calculate how much weight the crane could lift

AI Thread Summary
To calculate the weight a crane can lift, it's essential to consider the forces acting on the jib, including the mass and acceleration. The jib's different lengths create varying loads, particularly with a 16T counterweight on the shorter end. The mechanics of equilibrium must be analyzed to understand how these forces balance out. A sketch of the crane setup can aid in visualizing the problem and clarifying the calculations. Further assistance in mechanics is needed to resolve the confusion regarding the jib's dimensions and load distribution.
Clueless87
Messages
1
Reaction score
0
Homework Statement
Hi all, new to the forum so not sure if I’m filling this out correctly but I’m currently studying for an exam for my work and going through past papers. The question is;

The horizontal jib of a tower crane is 36m long, weights 3T and is mounted on a turntable 8m from the machinery end of the jib. The shorter end of the jib carries a 16T counter weight at its furthest extent.

i)Draw a sketch of the crane (east enough), and the loads exerted on it (not so easy…for me at least) (5 points)

ii) Calculate how much weight the crane could lift if the load were suspended at the far end of the crane’s jib and the jib is in equilibrium. (5 points)
Relevant Equations
I have no idea….
I think the first part is F = mxa? But given the jib is resting on a turntable at a certain distance, does that mean the longer and shorted ends experience different loads? Especially when the shorter end has a 16T counterweight?

The second part, I have no idea how to calculate this nor the equilibrium.

Any help would be greatly appreciated!
 
Physics news on Phys.org
Mechanics help?
Drop your car at the PF garage.
Welcome to PF!
 
Start by making a sketch and post it here!
 
Clueless87 said:
The horizontal jib of a tower crane is 36m long, weights 3T and is mounted on a turntable 8m from the machinery end of the jib. The shorter end of the jib carries a 16T counter weight at its furthest extent.
I can't make sense of that. 8<36/2, yet the other end is shorter? Should it be 28?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top