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August 2nd, 2019, 01:10 PM   #11
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Originally Posted by helpmeddddd View Post
The torque on the 2500N (far right of the diagram) should be 2500Nm ( I used the 1m for distance in the equation)

The torque for 1500N (IN the middle of diagram) I got 3000Nm(I used the 2m measurement)
No ... how far is the 1500N force from the $\color{red}{\text{pivot point}}$?

The torque for F is? ( I'd use 3m for distance)
the torque is F times 3.

Now finish it ...
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August 2nd, 2019, 02:04 PM   #12
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The wording of the question seems incorrect.

Stating "Without the force F the platform is just about to tip." is different from stating "Without the force F the platform would tip." The wording used suggests that any non-zero downward force F would suffice to prevent tipping. However, that wouldn't be the case if the distances and other forces shown are correct.

The suggested answer process would calculate the value of F such that the platform is precisely balanced with that force applied. However, this equilibrium wouldn't be stable.
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August 3rd, 2019, 06:15 PM   #13
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So for 1500N it's actually 1m so = 1500Nm

And what do you mean "F times 3"? Do I add up all the forces I've calculated?

Last edited by skipjack; August 4th, 2019 at 07:18 PM.
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August 4th, 2019, 02:37 PM   #14
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My working: T=fxd T(for force 2500N) = 2500x1= 2500Nm, (torque for force bridge downwards 1500N) =1500x1=1500Nm sum up both torques 2500-1500=1000Nm (solve for unknown weight F) Fx3 F=1000/3= 333N?
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August 5th, 2019, 02:21 PM   #15
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2500+1500=3f 4000=3f f=4000/3 f=1333N?

I'd really like to know how I'm going wrong. If I am. @skeeter

Last edited by helpmeddddd; August 5th, 2019 at 02:34 PM.
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August 6th, 2019, 12:04 AM   #16
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Your previous answer (333N) is correct.
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