# Static Equilibrium and Moment of Inertia problems

## Homework Statement

Prob1
The mechanical system of massless members is shown in the figure. Determine the relationship between two forces P1 and P2 to keep the system in equilibrium at position as showing in the figure.

Prob2.
Consider the shaded planar area with the y-axis as the axis of symmetry si shown in the figure. Determine the location (x, y) of the centroid and moment of inertia for the area about the x-axis.

The attempt at a solution

1)

M(A) = -P1*L+FBD*a = 0 -> FBD = P1*L/a

DEF is equilibrium :

Ex+P2=0
Ey+FBD=0

Ey/Ex = tan(30) = FBD/P2 = P1*L / a*P2

Is this right ?

Many Thanks.

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PhanthomJay
Homework Helper
Gold Member
Note that the slender members BD, DE, and DF are in compression, which is odd because this implies that these members are slender rods and not cords, which can't take compression. Looking at the sketch, I thought they were cords. But regardless, your solution for the (compressive) force in BD is good. But in looking at the slender members, should there also be a vertical force at F?

If there's a vertical force at F (eg Fx) it will be :

Ex+P2-Fx=0
Ey+FBD=0

I think I cant solve this.

PhanthomJay
Homework Helper
Gold Member
N0, you had the equation in the x direction correct the first time. If there is a vertical force at F, that's Fy, not Fx. From symmetry, what's the relation between Ey and Fy?

So you mean:

Ex+P2=0
Ey+Fy+FBD=0

where Ey=Fy

PhanthomJay
Homework Helper
Gold Member
Yes.

In Prob2, can I divide the shape to 2 rectangular :

The centroid of upper rectangular is located at (0,9/2) & the other one is (0,2a) -> The centroid of the shape is located at (0,43a/14) right. How can I find the moment of inertia for the area about the x-axis.

PhanthomJay