Torque diagram problem

1. May 7, 2003

Poutine

Hello all,
I'm having a problem with two of the problems I have to do, but I'm only going to put one here and the other one later if I can't figure
it out.

I made a quick diagram of the problem http://www.angelfire.com/nj4/angelus/prob.jpg [Broken]

The problem states:
In the figure one end of a uniform beam that weighs 214 N is attached to a wall with a hinge. The other end is supported by a wire.
a. find the tension of the wire
b. What is the horizontal component of the force on the hinge on the beam.
c. What is the vertical component of the force on the hinge on the beam.

Now from what my teacher told me I can place a pivot point at any place of my choosing than work with the torques with respect to that point. If I place the point at the hinge, then I find the sum of all torques which has to be equal to zero.

The problem is I'm not sure what numbers to use.

0 = -(214 N) * (L * cos(40)/2) + something.

(where L is the length of the beam)
I'm not sure what to put there. I know that if it were a horizontal beam it would be (L * T * sin(angle to the horizontal) but with the beam at an angle I'm not sure what to put.

Also how would I find the horizontal and vertical forces?

sum of all forces = 0 = horizontal component of beam + what?
sum = 0 = vertical component of beam + what?

Any help would be appreciated.

Last edited by a moderator: May 1, 2017
2. May 7, 2003

gnome

Don't forget that to compute the torque, you have to multiply each force by its moment arm...

Horizontal and vertical forces are ... horizontal and vertical forces (or the horizontal and vertical components of forces) , regardless of the angle of the object.

You have to set up 3 equations, one for the torques summing to zero, one for the horizontal forces summing to zero, and one for the vertical forces summing to zero, and solve them simultaneously.

You can help yourself by showing the forces on your diagram by putting arrows that indicate the point where each force acts on the object, and the direction of each force. That will make it easier to figure out the angles and trig functions to use. For example, ask yourself: where does the tension act and in what direction? where does the weight of the beam act and in what direction?

3. May 7, 2003

Poutine

Okay so for the Horizontal then it would have to be
- T(cos 60) + Horizontal com. of force = 0

The vertical force I'm less sure about.
Would it be
- T(sin 60) + Vertical comp at hinge - 214N = 0 (?)

But the torque one is still throwing me.

That moment arm of the string is going to L(cos 40) I believe and the moment arm of the beam is L (cos(40)/2)

But what into the formula for the torque?

0 = -(214 N) * L (cos(40)/2) + T(sin 20) * L(cos 40) (?)

4. May 7, 2003

gnome

Consider -- what is the significance of the + and - signs in that equation?

The moment arm for the string is a line perpendicular to the string, going through the hinge. I don't see where 40 works into that. As for the torque equation, remember that torque is just the force times the length of the moment arm. Think about that some more.