Gusset Plate Problem: Solving the Force on MN

[cwill53]

Homework Statement

Determine the force transmitted by the gusset late mnpqr to the member MN of the joint shown in the figure if the forces acting along the straight lines OA, OB, OC are $$P_{1}=P_{3}=141 lbs$$ and $$P_{2}= 100 lbs$$. The directions of these forces are shown in the figure.

Relevant Equations

$$F_{x}= \left \| \vec{F} \right \|cos\phi$$
$$F_{y}= \left \| \vec{F} \right \|sin\phi$$

Here is the image of the problem.

I arrived at the solution that the force on MN is in fact 100lbs due to P2 because the components P1 and P3 acting along MN are equal in magnitude but opposite in direction. What I don't understand is why the force is acting in the direction OPPOSITE to $$\vec{P_{2}}$$. That doesn't make sense to me.

 

[PhanthomJay]

You did not solve the problem correctly. You noted that the components of P1 and P3 cancel out when looking at the forces on the member along its length axis. correct. But what about the other components?

 

[cwill53]

You did not solve the problem correctly. You noted that the components of P1 and P3 cancel out when looking at the forces on the member along its length axis. correct. But what about the other components?

I see now. The other components of P1 and P3 add up to approximately 200lbs which opposes the force of P2 directly.