Recent content by petal5

Thanks again for all your help.Filling in for the equation in post #10 I get T2= 13321.274

But part (a) of the problem is: Find the forces exerted by the two cables.Then part (b) is:What are the torques exerted by each of the two cables and by the centre of mass of the beam about a pivot point at then lower endc of the beam.So I seem to need F2??

In my last post I said that F1=0 and then you replied that T1=0.Are both F1 and T1 equal to 0? Is the formula you mentioned above the formula for F2?

I've had another look at it and the angle and distance I posted above are incorrect.I've now come up with T1=r1f1sin65, T1=(0m)(F1)(0.9063).This is with using the lower end to take torques.However,I'm a bit confused as F1 turns out to be 0. For T2 I've gotten (1.6m)(F2)(0.9063) therefore T2=...

Thanks for your help,I really appreciate it! Do you mean to work out the following: T1=r1f1sinx , which would be T1=(0.4m)(F1)(sin25)

Sorry but what do you mean by 'talking moments'?

I'm thinking F1+F2=Mg but I'm not sure

I'd appreciate help with this problem as I don't know how to start it: A beam with a mass of 1500 kg and a length of 2.0 m hangs from two cables, such that it makes an angle of 65degrees with the vertical. Both of the cables are completely vertical; one is attached at the lower end of the...

The problem is: The angular speed of a propellor on a boat increases with constant acceleration from 50 rev/min to 150rev/min in 2.5s.What is the angular acceleration of the propellor? I'm not sure how to go about the problem.So far I've converted the 50rev/min and the 150rev/min to...

Thanks for all your help!So should my equation be: v^2/r=(0.050)(9.80)

I said 0.050g=v^2/r (taking v to be 61m/s)

I'm getting an answer of 74.42km for the radius.Does this sound about right to you?

em,in circular motion direction constantly changes.Centripetal acceleration is the resulting center directed acceleration. Do I need the formula: a=v^2/r ?

so far I've attempted to solve using v=r w