Calculating Support Forces on a Rigid Beam with Distributed Loads

[ice2morrow]

Homework Statement

A 5.00 m long rigid beam with a mass of 100.0 kg is supported at each end. An 80.0 kg worker stands on the beam, 4.00 m from the left support. 30.0 kg of brick are stacked on the beam 1.50 m from the left end of the beam. How much upward force does each support exert on the beam? Be specific by left support and right support?

Homework Equations

I'm not really sure

The Attempt at a Solution

I have no idea how to approach this problem.

 

[PhanthomJay]

Homework Statement

A 5.00 m long rigid beam with a mass of 100.0 kg is supported at each end. An 80.0 kg worker stands on the beam, 4.00 m from the left support. 30.0 kg of brick are stacked on the beam 1.50 m from the left end of the beam. How much upward force does each support exert on the beam? Be specific by left support and right support?

Homework Equations

I'm not really sure

The Attempt at a Solution

I have no idea how to approach this problem.

For a statically determinate problem like this, it doesn't matter whether the beam is rigid or flexible, because the reactions are independent of the beam deflection or rigidity. Just sum moments (torques) of the forces about the left end = 0, to get the right support reaction, then sum moments = 0 about the right end to get the left support reaction, then sum forces in the y direction = 0 to be sure you didn't make a math error.

 

[ice2morrow]

...Could I get a little more of a step by step? I'm not doing well in this course and have little to no understanding of this material.

 

[PhanthomJay]

Per Newton's laws, the sum of all forces for a body in equilibrium must equal 0, and the sum of all torques about any point must also equal zero. The 100 kg beam weighs 100g's or 980 N applied at the center of the beam, so it's torque about the left end is (980 N)(2.5 m) = 2450 N-m, clockwise. Now figure out the torques about the left end from the other loads, all will be clockwise, or plus, and don't forget the torque from the right support reaction (R_r)(5), counterclockwise, or minus; add 'em up and set them equal to zero , and solve for R_r. Continue.

 

[ice2morrow]

So left side is (980)(2.5) = 2450? Why is that? Shouldn't that be center? 2.5 is half the beam. And 980 is the torque on the center.

 

[PhanthomJay]

So the moment about[/color] the left side from the beam's weight force[/color] is (980)(2.5) = 2450?

Why is that?

The moment (torque) of a force about a point is equal to the product of that force times the perpendicular distance from the line of action of that force to the point. Here, the force is 980 N (W=mg =100 kg*9.8 m/sec^2 = 980 N) and the perpendicular distance is 2.5 m, because the resultant weight force acts at the cg (center) of the beam (assumed as a uniform beam with constant cross section).

And 980 is the torque on the center.

980 N is the resultant beam's weight FORCE applied at the center. You must now calculate the torque from the other forces, and solve per my previous response.