# Moments

• Sep 17th 2010, 10:58 AM
BabyMilo
Moments
A uniform beam AB of length 1.6m and weight 60N rests on two smooth supports at C and D where AC=BD=0.3m. A load is attached to A so that the supporting force at C is twice the supporting force at D. Find the magnitude of the load.
• Sep 17th 2010, 01:09 PM
emakarov
Write a system of two equations. First, the sum of all forces acting on the beam is 0. The forces are: the weight of the load, the beam's weight applied to the middle of the beam (these two are directed down), and two supporting forces at C and D (these are directed up).

The second equation says that the sum of torque (moment) is 0. Torque is calculated with respect to some pivot point, but in statics one can choose pivot point arbitrarily. So you can choose, e.g., the center of the beam or its left end. Then the torques of some of the four forces will try to rotate the beam clockwise and others counterclockwise; those torques come with opposite signs.
• Sep 17th 2010, 01:19 PM
BabyMilo
x=unknown force
Fc=force at C
Fd= force at D

x+60-Fc-Fd=0

not quite sure what you mean in the second equation, please show me.
I dont have a physics book and the maths book doesnt have any example.

thanks.
• Sep 18th 2010, 03:40 AM
emakarov
Since you are studying this topic, you must have some information concerning torque, e.g., from a textbook, from a lecture, or from some place online. Otherwise, it does not make sense to take on problems without first studying a general method of solving them.

Suppose we make the center of the beam the pivot point. Then the unknown force and Fd attempt to turn the beam counterclockwise, while Fc tries to turn it clockwise. The weight of the beam is applied directly to the pivot point; therefore, it is not turning the beam. (That's why it makes sense to choose the pivot point as the application point of one of the forces to eliminate this force from the equation.)

The torque of each force is the magnitude of the force times the distance between the pivot point and the application point of the force. The total torque (taking the signs into account) is zero.