What are the Internal Forces at Point C in a Distributed Load System?

[Isaac Reed]

Homework Statement

find internal normal, shear, and moment forces at point C, P=8kn

The Attempt at a Solution

for the most part I can solve this entire question but looking at the solution for it one of my equations is different and I cannot figure out why.
my equation for finding T(tension in cable)
ΣMa=0: 8(2.25)-T(.75/.96)=0
gives T=28.84kN

equation 2
ΣMa=0: -T(0.6)+8(2.25)=0 ←?
gives T=30.00kN

 

[paisiello2]

I don't see where your 1st equation comes from.

2nd equation looks right.

 

[Isaac Reed]

but why exactly the T(0.6) in 2nd equation

 

[paisiello2]

Isn't it the lever arm 0.5m+0.1m?

 

[Isaac Reed]

how does all of the tension in the cable get transmitted to the lever arm over the pulley?

 

[paisiello2]

Just as anything else. If the pulley is not rotating then the tension in the cable is constant for its entire length.

 

[PhanthomJay]

how does all of the tension in the cable get transmitted to the lever arm over the pulley?

You didn't respond to paisiello2 comment in post 2, but it appears you tried to treat this as a truss at the joint where load P is applied by isolating the joint and assuming all the vert applied load is taken by the cable. But some of it is taken in shear by the lower member. Trust your equilibrium equations when determining support reactions from externally applied loads.

 

[Isaac Reed]

Ahhhhhhh OK because the cable is only in the x direction after it goes over the pulley all of its y components are transferred to the beam. So if I finish my first equation I end up with.

ΣM=0: 8(2.25)-T(cos33.7)(2.25)+T(cos33.7)(1.5)=0
And I get T=30

 

[paisiello2]

Not sure what you did there but if you draw a proper free body diagram you will see directly that the 2nd equation in your OP is the most straight forward.

 

[Isaac Reed]

It is clearly the most straiforward. But how is the component of the tension at the end of the beam not included in equation?

 

[paisiello2]

Draw a free body diagram. The rope is only cut at one location. The rest is internal.