Forces and torque on a cable/beam

[MitsuShai]

Homework Statement

Directions: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/directions.jpg
Question: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q2.jpg

Homework Equations

f=ma
torque= rFsin(theta) or = rF

The Attempt at a Solution

I am really bad at these types of problems, so I started drawing all the forces: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/excrques.jpg

good so far or not?

 

[Delphi51]

The 66 degrees between the cable and the beam is not correct. Should be 66-39.

I would concentrate on the tip of the beam, drawing the 3 forces that act there. They must, of course, add up to zero. Unfortunately, I think two of the forces are unknowns so you are going to need a second equation. Torque?

 

[MitsuShai]

The 66 degrees between the cable and the beam is not correct. Should be 66-39.

I would concentrate on the tip of the beam, drawing the 3 forces that act there. They must, of course, add up to zero. Unfortunately, I think two of the forces are unknowns so you are going to need a second equation. Torque?

ok I corrected the picture so now for the question

a) the angle between the cable and the beam is 37 degrees and the angle
and the angle between the cable and the beam is the angle is where the pink arrow and star are in the picture, right? and that angle would be 14 degrees

b) Fx= Fwall - TCos(39) = 0
Fy= F(n of beam) - F(w beam) - TSin(39) - F(load) = 0

c) the reference point is where the purple star is, so
torque= -(L)F(weight of beam) + (L)F(normal of beam) - (L)F(w load) - (L)Tsin(39) +(L)F(n load) + T(load)=0

good so far or not?

 

[Delphi51]

a) the angle between the cable and the beam is 37 degrees and the angle
and the angle between the cable and the beam is the angle is where the pink arrow and star are in the picture, right? and that angle would be 14 degrees

I'm still getting 66-39 = 27 for the angle between the cable and the beam, where you have marked 37. The angle between the beam and horizontal is 66. From that you must subtract the 39 degrees between the cable and horizontal.

b) Fx= Fwall - TCos(39) = 0
Fy= F(n of beam) - F(w beam) - TSin(39) - F(load) = 0

I'm not following this. What is Fwall? If you mean the force with which the wall pulls on the rope, wouldn't that be T?

c) the reference point is where the purple star is, so
torque= -(L)F(weight of beam) + (L)F(normal of beam) - (L)F(w load) - (L)Tsin(39) +(L)F(n load) + T(load)

Okay, finding torques about the star point. Assuming the beam is free to turn about this point, so the total torque is zero. The weight of the beam is at an angle to the beam length L, and acts at its center of mass, so I think that torque is something like
0.5*L*cos(66)*(weight of beam).
I don't know what you mean by F . . .

 

[MitsuShai]

I'm still getting 66-39 = 27 for the angle between the cable and the beam, where you have marked 37. The angle between the beam and horizontal is 66. From that you must subtract the 39 degrees between the cable and horizontal.

I'm not following this. What is Fwall? If you mean the force with which the wall pulls on the rope, wouldn't that be T?

Okay, finding torques about the star point. Assuming the beam is free to turn about this point, so the total torque is zero. The weight of the beam is at an angle to the beam length L, and acts at its center of mass, so I think that torque is something like
0.5*L*cos(66)*(weight of beam).
I don't know what you mean by F . . .

a) right sorry it is 27 degrees
and the angle between the cable and the beam is 39 + 90 = 129 degrees

b) Fwall is the force the wall exerts on the beam and I used T for the tension of the cable
Fx= Fwall - TCos(129) = 0
Fy= F(n of beam) - F(w beam) - TSin(39) - F(load) = ma

c) F is just the force from the equation torque= rF

[/QUOTE] Okay, finding torques about the star point. Assuming the beam is free to turn about this point, so the total torque is zero. The weight of the beam is at an angle to the beam length L, and acts at its center of mass, so I think that torque is something like
0.5*L*cos(66)*(weight of beam).
I don't know what you mean by F . . .[/QUOTE]

wait a minute where did the 66 degrees come from, I think I'm looking at this wrong
I was thinking of something like this:
http://i324.photobucket.com/albums/k327/ProtoGirlEXE/closeup.jpg

 

[MitsuShai]

wait nevermind the angle would still be 39 degrees not 129

 

[Delphi51]

Fx= Fwall - TCos(129) = 0
Fy= F(n of beam) - F(w beam) - TSin(39) - F(load) = ma

We normally sum the forces on a point but these forces act far apart on the ends of the beam. I don't think we can say they total zero. The wall will push vertically as well as horizontally on the beam.

It seems to me the sum of the forces on the tip of the beam would be zero, and the equation would be useful. Also, the torques about the beam's pivot point (if it is free to turn where it is attached to the wall).

 

[MitsuShai]

We normally sum the forces on a point but these forces act far apart on the ends of the beam. I don't think we can say they total zero. The wall will push vertically as well as horizontally on the beam.

It seems to me the sum of the forces on the tip of the beam would be zero, and the equation would be useful. Also, the torques about the beam's pivot point (if it is free to turn where it is attached to the wall).

ok I got the hang of this problem now, thank you so much! :)