Lifting Device Question for anyone who can help me (1 Viewer)

[scorch62]

Hey guys been thrown completely by this lifting devices question. Can anyone explain or do this step by step?

View attachment 20666

Thanks.

 

[iEdd]

Start by drawing a free body diagram. You have one force 4900N down, point A is in equilibrium. Assuming the crane cable is held tightly and can't move, what force do you need to pull @ 35° to pull it at an angle of 10°?

 

[notmyrealname]

Hi Scorch,

Following on from iEdd's reply, draw a Free Body Diagram of the system at point A and then resolve forces using a force triangle. (google is your friend if you don't know what the force triangle is.)

They've used g=10 m.s^-2 for this example, and the correct answer is (e.)

They've mucked up their units. The answer's in Newtons, not in kN

Cheers

 

[scorch62]

Yeah I got it now.
Answer is 5792.28N

 

[notmyrealname]

Yeah I got it now.
Answer is 5792.28N

Hi Scorch

Take II at a response. [edit: I use firefox/no-script and I've lost a few attempts. Sorry for brief summary.]


From the force triangle:

Force in the rope being pulled by the person as identified:

5000/sin 45° = F(rope) x sin 10°

F(rope) = 5000/sin45° x sin10°

= 1227.9N

= 1230 N (rounded to three significant figures.)




You've inadvertently, but accurately, calculated the tension in the supporting cable at 10 degrees to the vertical.

Cheers,

 

[scorch62]

Oh damn. You are right
Yeah I see where I've gone wrong now. It's what happens when you don't label your vectors when you draw.