- #36
Doc Al
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You treat the tension in problem 2 exactly as you did in problem 1.
Yes.ssj said:Right to in problem 1 we basically made it perpendicular correct ?
If you're talking about problem 1, that should be:So for now let's call the length L meaning that 200*L=200*T*sin30,
Using the corrected equation, we get: T = 200/sin30 = 400 N. The L drops out and has no bearing on the answer for tension. If the beam were twice as long, you'd still get the same tension.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.
The right side is correct, but the left is not. Don't guess! You have two forces producing clockwise torques--add up the torque from each. You should know the torque from each one, since they are the same forces as appeared earlier.ssj said:My guess is something like 400*L/4=L*T*sin30
Doc Al said:The right side is correct, but the left is not. Don't guess! You have two forces producing clockwise torques--add up the torque from each. You should know the torque from each one, since they are the same forces as appeared earlier.
Those were the tensions, not the torques. The clockwise torque from problem 1 was 200*L; from problem 2 it was 200*L/2.ssj said:So the torque from question 1 was 400N and question 2 was 200N meaning we have
600*L/2=L*T*sin30.
Doc Al said:Those were the tensions, not the torques. The clockwise torque from problem 1 was 200*L; from problem 2 it was 200*L/2.
No:ssj said:Right this makes it 400*L/3=L*T*Sin30
Doc Al said:No:
200*L + 200*L/2 = 200*(3L/2)
This is not the equilibrium equation, this is just the sum of the clockwise torques (see post #43): 200*L + 200*(L/2) = 200*(3L/2) = 300*Lssj said:How did you get "3L/2" on the right hand side and why isit L/2 on the left side ? Furthure more once I calcualted this euqation L=200 where do I go from here?
Doc Al said:This is not the equilibrium equation, this is just the sum of the clockwise torques (see post #43): 200*L + 200*(L/2) = 200*(3L/2) = 300*L
Now set this sum of clockwise torques equal to the counter-clockwise 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.
L + L/2 = 3L/2ssj said:For the sum of the clockwise torques I understand that 200*L+200(L/2) but I don't understand 200*(3L/2) where did the "3L" come from ?
Doc Al said:L + L/2 = 3L/2
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.
The left-hand side (which represents clockwise torques) is correct--it's the same as the last problem. But the right-hand side is not correct. The angle is not the only thing that changes: the distance to the pivot point also changes--it's not L anymore.ssj said:I would think the answer is 300*L=L*T*Sin50 .
Doc Al said:The left-hand side (which represents clockwise torques) is correct--it's the same as the last problem. But the right-hand side is not correct. The angle is not the only thing that changes: the distance to the pivot point also changes--it's not L anymore.
You now have the correct distance on the right-hand side (L/2), but what happened to the angle?ssj said:So this would make it 300*L=L/2*T*Sin30 ?
Doc Al said:You now have the correct distance on the right-hand side (L/2), but what happened to the angle?
Good! (I thought so.) Now you can solve for the tension--once again, the actual value of L doesn't matter.ssj said:Sorry about that I meant Sin50.
Right so the equation now becomes 300*L=L/2*T*Sin50.Doc Al said:Good! (I thought so.)
Yep.ssj said:Right so the equation now becomes 300*L=L/2*T*Sin50.
Doc Al said:Yep.
Correct. But realize you don't have to "replace" L with any number--the L just drops out:ssj said:So if I replace L with 2 I get T=600/sin50 meaning T=783.
Another equilibrium problem. What are the conditions for equilibrium? Forces must add to zero and torques about any point must add to zero. Label the vertical support forces (at each end of the bridge) F1 & F2. Hint: A good point to use as a pivot is either end of the bridge.Now I am stuck with the following question ...
Doc Al said:Another equilibrium problem. What are the conditions for equilibrium? Forces must add to zero and torques about any point must add to zero. Label the vertical support forces (at each end of the bridge) F1 & F2. Hint: A good point to use as a pivot is either end of the bridge.
No. F1 is the upward force of the support on the left end of the bridge, F2 is the upward force on the right end of the bridge.ssj said:Alright so if I lable the force between 4kN and 10kN F1 and from 10kN to 8kN F2 and place an pivot at the left hand side of the bridge would this be correct and what would the equation be for the clockwise rotation= counter clockwise rotation ?
No. There are three forces (each at a different distance from the pivot) that create clockwise torques. And the force F2, which creates the counter-clockwise torque, is how far from the pivot?ssj said:Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?
Doc Al said:No. There are three forces (each at a different distance from the pivot) that create clockwise torques. And the force F2, which creates the counter-clockwise torque, is how far from the pivot?
I encourage you to study the examples discussed on the webpage I linked in post #37.
Doc Al said:No, the distance of F2 from the left pivot point is: 7m + 3.5m + 5.25m + 5.25m = 21m. (You didn't even have to do that addition: the problem states that the bridge supports are 21m apart.)