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Melawrghk
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This is just a problem from my midterm, and I was wondering if I did it correctly
Weight of block B is 15kN, weight of block A is 25kN. Find force P that has to be applied to keep the system in equilibrium in tension in the cable
[tex]\sum[/tex]Fx=0
[tex]\sum[/tex]Fy=0
So first I figured I'd draw a FBD around block B (FBD1 on the image). I also decided I would use a different axis (x' and y'). From this I was able to write the equilibrium equations:
[tex]\sum[/tex]Fx=3T-15*cos(60)=0, from which T=2.5kN
(I also wrote the Fy equation, but I won't post it because it serves no real point)
Next, I drew a FBD around block A (FBD2 on the image). I used a different axis once again. And I got:
[tex]\sum[/tex]Fx=-25*cos(60)+P(cos30)+2.5kN=0, from which P=11.55kN
Is that correct? It makes sense in my head, but then again that wouldn't be the first time my gut feeling is wrong.Thanks!
Homework Statement
Weight of block B is 15kN, weight of block A is 25kN. Find force P that has to be applied to keep the system in equilibrium in tension in the cable
Homework Equations
[tex]\sum[/tex]Fx=0
[tex]\sum[/tex]Fy=0
The Attempt at a Solution
So first I figured I'd draw a FBD around block B (FBD1 on the image). I also decided I would use a different axis (x' and y'). From this I was able to write the equilibrium equations:
[tex]\sum[/tex]Fx=3T-15*cos(60)=0, from which T=2.5kN
(I also wrote the Fy equation, but I won't post it because it serves no real point)
Next, I drew a FBD around block A (FBD2 on the image). I used a different axis once again. And I got:
[tex]\sum[/tex]Fx=-25*cos(60)+P(cos30)+2.5kN=0, from which P=11.55kN
Is that correct? It makes sense in my head, but then again that wouldn't be the first time my gut feeling is wrong.Thanks!