0.14 Phy1110: force -- moments, torque, couple, and equilibrium  (Page 9/13)

Insert a hypothetical spring scale in each rope

Now cut each rope somewhere along its length and insert something that is meant to replicate a spring scale in each rope. Assume that the weights of theropes, the spring scales, and the bar are negligible. Also assume that the extension of the spring scales caused by loading is negligible. (In other words,the bar will remain horizontal even when downward forces are applied.)

Each spring scale will exert an upward force on the bar and will, at the same time indicate the amount of force with which that end of thebar is pulling down on the rope that supports it.

The bar is in equilibrium

By default, the bar is in equilibrium. By that I mean that the bar doesn't accelerate in any direction, nor does it rotate about any axis. Furthermore, theweight being registered on each spring scale is negligible.

Label points and distances on your drawing

Use your Braille labeler to label the left end of the bar A and the right end of the bar B. Use pins to subdivide the bar into ten units.

Now assume that you will hang a 10 N weight somewhere on the bar between the ropes. Label that point C. Label the distancefrom C to A as a, and label the distance from C to B as b. Label the weight as W.

Add the weight to the bar

The situation changes when you add the 10 N weight to the bar. Each spring scale now exerts an upward force on its end of the bar that is no longer negligible. Label the upward force onthe left end P and label the upward force on the right end Q.

Is the bar in equilibrium

I believe that most of you will know, based on experience with the trapeze bar on the playground, that the bar will still be in equilibrium. When you hungquietly on a trapeze bar as a child, it didn't accelerate off in some direction, nor did it rotate about any axis. So, we know from experience thatthe bar is still in equilibrium.

Three forces

There are now three forces being exerted on the bar: P, Q, and W. We know from earlier modules that the vector sum of those three forces must be zero in orderfor the bar to be in equilibrium. This system is simple enough that we can perform the vector sum in our heads. We don't need to draw a vector diagram. We see that

P + Q - W = 0

Adding W to both sides of the equation gives us:

P + Q = W ( eq. a1 )

The values for P and Q

What are the values for P and Q? We know that wherever we place the weight along the horizontal length of the bar, the bar will continue to be inequilibrium. Therefore, the sum of P and Q must be equal to W regardless of the position of W (so long as the position of W is between the ropes and the ropesdon't break).

Compute the torque about C

Assume that positive coordinates are to the right with the origin at the left end of the bar. Computing the torqueabout the point C gives us:

(C-A)*(P) + (C-B)*(Q) = 0

Let's put some numbers in the problem now.

• Let C = 2
• Let A = 0
• Let B = 10
• Let W = 10

Substituting numbers in the above equation gives us:

(2)*(P) + (-8)*(Q) = 0

Simplify and rearrange terms