# Rigid Body Equillibrium

## Homework Statement

As shown, a uniform beam of length l = 5.90 and 48.2 lb is attached to a wall with a pin at point B. A cable attached at point A supports the beam. The beam supports a distributed weight = 21.0 lb/ft. If the support cable can sustain a maximum tension of 300 lb , what is the maximum value for W1 ? Under this maximum weight, what is FBy , the vertical component of the support's reaction force at point B?

## Homework Equations

$\Sigma$ F = 0
$\Sigma$ M = 0

## The Attempt at a Solution

I began by trying to draw a free-body diagram with everything I know put in place. I then came up with 3 equations: The sum of x-axis forces, the sum of y-axis forces, and the sum of moments at point B.

(1) $\Sigma$Fx: 300 cos(60o)+ Bx= 0
(2) $\Sigma$Fy: 300 sin(60o)- 48.2 - 123.9 - W1 - By= 0
(3)( this is my problem equation)$\Sigma$M: -300cos(60o )(5.9ft) - (127.1)(2.95) - W1(0) - By(0) = 0

The problem I can't quite figure out is that if the unknowns FBy and W1 are applied at B, they create no moments, and that equation becomes useless for solving my system. So, I'm left with an indeterminate system of 2 equations and 3 unknowns. I've attached the Figure referenced in the problem statement and a MS Paint version of my FBD. I just need a nudge in the right direction I think.

I also replaced the distributed load with an equivalent Force of 123.9 lb directed at the center of the beam.

Addendum: I also tried with By oriented upward, AND I tried maybe summing the moments at A instead. I still get an indeterminate answer.
Thank you in advance for your help

Last edited:

## Answers and Replies

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
You have errors in your calculations:

1. In equation 1, the term 300 cos (60) represents the vertical component of the line tension, not the horizontal. The location of the angle makes a difference in when to use sin/cos for the horiz/vert component calculation. Likewise, your horizontal line component needs correction in your equation 2.

2. w1 and w2 are not concentrated forces, but instead represent distributed loads in lbs/ft. Thus in your moment equation 3) the moment of w1 which you have written is incorrect. Given that w2 = 21 lbs/ft, the problem is asking you to calculate how large w1 can be in lbs/ft such that the tension in the line does not exceed 300 lbs. Given that w2 = 21 lbs/ft, the load on the beam is:
21*5.9 = 123.9 lbs PLUS (w1 - 21)*5.9.
Thus, your equation 2 needs correction.

3. The moment of the distributed load uses 127.1*2.95 instead of 123.9*2.95

4. Likewise, an additional moment term is required for the triangular distributed load which has a value of (w1 - 21) lbs/ft at the wall.

Make these corrections and see if you can solve for w1.

• 1 person
So am I then treating W2 and a rectangular load and w1 as the remaining triangular portion in the diagram? I suppose that would produce two equivalent loads in the middle of the bridge (W1eq and W2eq) which would give me a W1 moment term to solve for)

Thank you so much for your help. I did the (in hindsight) dumbest thing possible by trying to take Statics as a 5-week summer course. 5 chapters in 12 days with no lecture.

I really appreciate your assistance!

Cheers

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
Yes. In order to avoid confusion, most texts use capital letters like W and P to denote concentrated loads. For distributed loads, lower case letters are used, which in this case would be w1 and w2.

• 1 person
You have errors in your calculations:
Given that w2 = 21 lbs/ft, the load on the beam is:
21*5.9 = 123.9 lbs PLUS (w1 - 21)*5.9.
Thus, your equation 2 needs correction.

That makes it seem like w2 is completely cancelled out by w1

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
The trapezoidal load distribution is 21 lbs/ft at the left end of the beam and w1 lbs/ft at the right end. Presumably w1 > w2, therefore w2 + x = w1. x = w1 - w2, where x represents the amount by which w1 is greater than w2.

The trapezoidal distributed load can be split into a constant load distribution plus a triangular load distribution. The constant load distribution is w2 over the length of the beam, and the triangular load distribution is 0 at the LH end of the beam and w1 - w2 at the RH side of the beam. The center of the constant distribution is L/2 from the RH end of the beam and the triangular distribution has a center L/3 from the RH end.

In looking over your moment equation again, I think you forgot to add in the moment due to the weight of the beam, or 48.2 * 2.95.