# Early Statics Question

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1. Sep 18, 2016

### tyrostoken

1. The problem statement, all variables and given/known data

Given: A 300 kg walkway supported by two rollers(at points A and B) with center of mass at G.
Find: The tension T in the horizontal cable attached to the cleat at point B and the force under the roller at A.

2. Relevant equations
∑M = 0
ΣFx = 0
ΣFy = 0

3. The attempt at a solution
Free body diagram of the bridge has Normal force vertically upwards with point of application A(Na) , Weight(W) vertically downwards at G, Normal force perpendicular to walkway directed away from water(Nb), and Tension(T) horizontal to the right at point B.

∑Mb = 0 (moment at point B)
0 = +W(4cos30°) - Na(8cos30°)
Na = (W(4cos30°))/(8cos30°)
Na = 1472 N

So far this answer is known to be correct for Na.

ΣFx = 0
0 = T - Nbcos30°
T = Nbcos30°
ΣFy = 0
0 = Na - W + Nbsin30°
Nbsin30° = W - Na
Nb = (W - Na) / sin30°

Plugging in the known values and Na that we solved for yields the incorrect answer. Why can't I add up the components of the force vectors or what am I doing incorrectly?

2. Sep 18, 2016

### PhanthomJay

Check your trig again when determining the x and y components of the normal force at B.

3. Sep 18, 2016

### Staff: Mentor

Check this.

4. Sep 18, 2016

### tyrostoken

OK the horizontal component of Nb should be Nbcos60° and the vertical component of Nb should be Nbsin60°. I think I was tired and messed up the trig. Thanks for the help.