Center of mass and Tension

1. Feb 19, 2010

y3ahright

1. The problem statement, all variables and given/known data
A man is standing 1 meter in from the left of a 4 meter long beam. The beam is held up by two wires perpendicular to the beam. The beam is 750 N and the man is 100kg what are the tensions of the two wires.

2. Relevant equations

F = ma

Torque = F x d

3. The attempt at a solution
I know that because the beam is not accelerating the sum of the forces is 0

so Fy = -750 - 980 + T1 + T2 = 0

T1+T2 = 1730

T2 = 1730 - T1

Torque (I put my pivot point at the first tension wire making it 0)

Torue = F x d = T2 * 4Sin(90) - 1730

T2 *4 = 1730

T2 = 423.5

1730 - 423.5 = 1306.5 = T1

Now the book says 1130 T1 and 610 T2

I know for sure i'm missing something with center of mass im just not sure what.

2. Feb 19, 2010

PhanthomJay

yes
what is this? Are you taking the torque from the man and beam weights as (1730 N)*(1 m)? The man's weight acts at 1 m, but the beam's weight does not act there.
You are not summing moments correctly. Where does the beam's weight of 980 N act on the beam ( how far from the left end?)?

3. Feb 19, 2010

y3ahright

so i should use Torque = -980 N (man)*1 meter (away from my pivot point of the left end of the beam) * sin(90) making it -980

and then add that with the -750N (beam) * 2meters (center of mass from the beam because it is uniform so it's l/2 or 4/2)

and that would leave T2* 4 meters because it's on the other end of the beam?

4. Feb 19, 2010

PhanthomJay

yes, good, add em up and solve for T2, then T1 will follow. As a check on your work, try summing moments about T2, the right end. The number you get for T1 should agree.

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