Equilibrium on a Beam: Determining Forces at Different Points

[tigerwoods99]

Homework Statement

A 20.0 m long uniform beam weighing 670 N is supported on walls A and B, as shown in Fig. 9-77.
http://www.webassign.net/giancoli5/9_77.gif

(a) Find the maximum weight a person can be to walk to the extreme end D without tipping the beam.
(b) Find the forces that the walls A and B exert on the beam when the same person is standing at point D.
(c) Find the forces that the walls A and B exert on the beam when the same person is standing at a point 2.0 m to the right of B.
(d) Find the forces that the walls A and B exert on the beam when the same person is standing 2.0 m to the right of A.

I figured out A + B but not C and D.

Homework Equations

sigma Fy = 0
sigma FTorque = 0

The Attempt at a Solution

FA = beam A
FB= beam B
mg= mass of beam * 9.8
FC= the person


a) 670N
b) 0 N from wall A and 1340N from wall B
C) using FA as the pivot: 0FAsin90 + 12FBsin90 -8.5mgsin90 - 15FCsin90 = 0
12FB - 15745 = 0
FB = 1312.08 WHICH IS NOT CORRECT.
D)

 

[PhanthomJay]

You have your dimensions wrong. If summing torques about A, the distance fron the cg of the beam to A, for example, is 7 feet. Also, you don't need m and g, since the force units of weight are already given.

 

[ideasrule]

b) 0 N from wall A and 1340N from wall B
C) using FA as the pivot: 0FAsin90 + 12FBsin90 -8.5mgsin90 - 15FCsin90 = 0
12FB - 15745 = 0
FB = 1312.08 WHICH IS NOT CORRECT.

What's FC? Where did you factor in the weight of the beam or the weight of the person?

 

[tigerwoods99]

What's FC? Where did you factor in the weight of the beam or the weight of the person?

FC = the force of the person on the beam as mentioned in the question.

The weight of the beam is mg.

Thanks.

 

[tigerwoods99]

You have your dimensions wrong. If summing torques about A, the distance fron the cg of the beam to A, for example, is 7 feet. Also, you don't need m and g, since the force units of weight are already given.

Thanks jay. Your correct the mg is 7 feet to the right of the FA, or beam A. When i wrote mg i was referring to the Force units.

How do i go from here?

Thanks!

 

[PhanthomJay]

Thanks jay. Your correct the mg is 7 feet to the right of the FA, or beam A. When i wrote mg i was referring to the Force units.

How do i go from here?

Thanks!

You do have parts 'a' and 'b' correct. For part 'c', try summing torques about point A, and solve for the unknown reaction force at B. Then solve for the unknown reaction force at A by summing forces in the y direction = 0. As a check, sum torques about point B = 0 to determine the reaction force at A. You can solve part 'd' in a similar fashion.

 

[tigerwoods99]

You do have parts 'a' and 'b' correct. For part 'c', try summing torques about point A, and solve for the unknown reaction force at B. Then solve for the unknown reaction force at A by summing forces in the y direction = 0. As a check, sum torques about point B = 0 to determine the reaction force at A. You can solve part 'd' in a similar fashion.

Thanks Phanthom Jay!

Could you take a look at this: https://www.physicsforums.com/showthread.php?p=2544626

Thanks again.