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Statics Homework help  find tension 
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#1
Oct1609, 07:25 PM

P: 7

[b]1. In Figure P4.111 determine the tensions in cables BC and BE. Neglect the weights of all members and assume that the support at A is a ballandsocket joint. The 5200N force has no x component.
(picture is attached below) [b]2. [tex]\Sigma[/tex] F_{x} = 0 [tex]\Sigma[/tex] F_{y} = 0 [tex]\Sigma[/tex] F_{z} = 0 [tex]\Sigma[/tex] M_{D} = 0 3. The attempt at a solution Okay so I figured out the angles of the cables and did the sum of the forces. I had trouble figuring out the sum of the moments. Here is what I have so far: [tex]\Sigma[/tex] F_{x} = 0 > Ax + T1*cos(26.6) = 0 [tex]\Sigma[/tex] F_{y} = 0 > Ay + T1*sin(26.6) + T2*cos(18.4) + 5200*cos(22.6)= 0 [tex]\Sigma[/tex] F_{z} = 0 > Az + T2*sin(18.4) + 5200*sin(22.6) = 0 T1 is the cable from B to E T2 is the cable from B to C Anything you guys can do to help would be much appreciated, thanks! 


#2
Oct1709, 10:44 AM

P: 4

It's much easier to do this geometrically:
Step 1  Find the y and z components of the 5200N force. Step 2  Multiply these by 12m (to get the torque) and divide by 6 m to find the respective forces that the cables need to provide equilibrium. Step 3  now you have four vectors: BE, BC, F(z), F(y). Step 4  Only BC can handle the vertical force (F(z)), so you know that the vertical component of BC = F(z). Step 5  Find the horizontal component of F(z). Step 6  Now you have your final equation, namely: F(y)  BC(y) = Horizontal component of BE. Step 7  Use Pythagoras' Theorem or Trig to find BC and BE Step 8  Be happy. Peace out! 


#3
Oct1709, 04:44 PM

P: 7

Is there another way to do this?
because my professor wanted us to use the sum of forces/moments. 


#4
Oct1809, 07:29 PM

P: 7

Statics Homework help  find tension
Hey so I figured out the majority of the problem.
Im just having trouble breaking the three forces into their components. (mostly with the angle) Can someone please tell me if this is correct. T1 = BE = T1cos(26.6)[tex]\hat{i}[/tex] + T2sin(26.6)[tex]\hat{j}[/tex] T2 = BC = T2cos(18.4)[tex]\hat{j}[/tex] + T2sin(18.4)[tex]\hat{k}[/tex] F = 5200cos(22.6)[tex]\hat{j}[/tex] + 5200sin(22.6)[tex]\hat{k}[/tex] pleeeease help. thanks! 


#5
Oct1909, 08:17 AM

P: 4

I'm too lazy to check all of your angles, but it looks reasonable. Just double check your SohCahToa and you should be fine.



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