How do I find cord tension in Equilibrium problem?

[pokeefer]

Homework Statement

Im doing Physics 12 with equilibrium

The diagram shows a beam 3.2 m long with masses hanging from it. The mass of the beam is negligible in comparison with the other masses.

Diagram:
http://i51.tinypic.com/jjb449.jpg

a.) If the beam is in equilibrium, what is the force of tension in the cable?

b.) If the beam is in equilibrium with the 150kg mass 3.2 m from the 90 kg mass, how far is the 90kg mass from the cable?

c.) If the force of tension in the cable is 2352 N and the cable is attached 2.2 m from the right end of the beam, what is the net torque on the beam about the right end?

d.) If the force of tension in the cable is 2352 N and the cable is attached 2.2 m from the right end of the beam, what is the torque about the point where the cable is attached to the beam?

Homework Equations

Net Torque = 0
Net Fx = 0
Net Fy = 0

Torque = Force x length
Torque = mass x gravity x length

The Attempt at a Solution

There were many similar examples but not in this design. So I am unsure if I am doing it right.

If you could at least answer question part a

I would greatly appreciate it

Thanks

 

[PhanthomJay]

For part a, why not try it yourself, using your third relevant equation? When a beam is in equilibrium, all the forces must sum to zero. There are 3 forces acting on the beam; identify the magnitudes and directions of the known forces, and solve for the unknown tension force in the cable.

 

[pokeefer]

For part a, why not try it yourself, using your third relevant equation? When a beam is in equilibrium, all the forces must sum to zero. There are 3 forces acting on the beam; identify the magnitudes and directions of the known forces, and solve for the unknown tension force in the cable.

Ok thanks,

Now I am just halfway through part c and a bit stuck.

And I don't know how I would go about part d :P

 

[PhanthomJay]

For part c, just sum torques of all forces about the given point, per your 4th relevant equation. Watch plus and minus signs (if you choose counterclockwise as plus , then clockwise is minus). The beam will not be in equilibrium, since there will be a net torque acting about the given point (that is, the net torque will not be 0).

Do the same for part d...what do you get?

 

[pokeefer]

For part C,

This is what I have so far. Since the mass of the beam is negligible:

2352N = F1 + F2

(2352N)(2.2m) + (2352N)(1m) = F1(3.2m)I'm trying to solve for Force 2 which is at the right end.

 

[PhanthomJay]

No, when you sum moments, it must be about a given point or chosen point, with the same point being used for all forces. You are choosing different points. Note that F1 and F2 are given as the weights of the masses, so you don't need to solve for them.

 

[pokeefer]

No, when you sum moments, it must be about a given point or chosen point, with the same point being usor all fed forces. You are choosing different points. Note that F1 and F2 are given as the weights of the masses, so you don't need to solve for them.

I'm really confused now.

How would I go about calculating the net torque?

What value does it want me to give it?

Aren't I supposed to solve for F2, isn't F2 the torque?

I'm sooo confused :(

Is it Net T = -(1m)(2352N) + (3.2m)F2 = 0

OR

Net T = -(1m)(2352N) - (2.2m)(2352N) + (3.2m)F2 = 0

 

[pokeefer]

This is depressing me so much.

I'm on my own and I have no teachers to help me at all. I'm doing online school and I'm spending hours trying to figure out what iv'e done wrong. :(

I have a very limited time now to finish this homework assignment and get through all the other assignments before taking a test and I am already failing the first lesson. :(
..................................................

 

[PhanthomJay]

Suppose you had a beam 5 m long and there was a force of 20N acting down at the left end of the beam , a force of 30 N acting up at 3 m from the right end of the beam, and a force of 10 N acting down at the right end of the beam. Now you are asked to calculate the net torque about the right end of the beam.

Net Torque = 20(5) - 30(3) + 10(0) = 10 N*m , counterclockwise.

Does that help?

Is it Net T = -(1m)(2352N) + (3.2m)F2 = 0

F2 is given as 90(9.8) = 882 N. So solve for Net Torque. The Net torque is not 0, because for part c , the beam is not in equilibrium.