
#37
Sep2807, 03:29 PM

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P: 40,882

You might want to study these examples: Torque Equilibrium Examples




#38
Sep2807, 03:34 PM

P: 36

Right to in problem 1 we basically made it perpendicular correct ? So for now lets call the lenght L meaning that 200*L=200*T*sin30, sin30 because that will give use the hypotenues which is T but the problem with the formula is that it gives the same answer as we got in question 1 meaning we must divide sin30 on each side meaning 200*L/sin30= 400 and once we find out the L we will get an new answer perhaps.




#39
Sep2807, 03:57 PM

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P: 40,882

200*L = L*T*sin30 If you are talking about problem 2, the left side of the equation is different. The distance is not L, but L/2. The right side remains the same, since the cable has not moved and the angle is the same. Rewrite the equation for problem 2 and solve for the tension. Again, the length of the beam (L) will drop out. 



#40
Sep2807, 05:22 PM

P: 36

Yes I think i understand 200*L/2=L*T*sin30
Which becomes 200/sin30/2 400/2=T 200=T 



#41
Sep2807, 05:30 PM

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P: 40,882

Good! In the second problem it takes just half the tension to support the beam.




#43
Sep2807, 06:00 PM

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#44
Sep2807, 06:05 PM

P: 36

600*L/2=L*T*sin30. 



#45
Sep2807, 06:24 PM

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#46
Sep2807, 06:34 PM

P: 36





#48
Sep2907, 04:28 AM

P: 36





#49
Sep2907, 06:23 AM

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P: 40,882

Now set this sum of clockwise torques equal to the counterclockwise torque due to the tension to the get the equilibrium equation: 300*L = L*T*sin30 Now you can solve for the tension in problem 3. 



#50
Sep2907, 07:11 AM

P: 36





#51
Sep2907, 07:45 AM

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P: 40,882

You can also write that as: (3/2)*L You can also do this: 200*L + 200*(L/2) = 200*L + 100*L = 300*L Please convince yourself that these are equivalent. 



#52
Sep2907, 11:44 AM

P: 36




#53
Sep2907, 11:48 AM

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P: 40,882





#54
Sep2907, 11:51 AM

P: 36




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