Rhys_M
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Here is a question from the Oxford OCR student book about couples and torques:

Figure 4 shows a rod of length 50cm with a disc of radius 10cm fixed at its centre. Two forces, both 30N, are applied normal to the rod at each end. Calculate the torque produced by the pair of 30N forces, and then the minimum tension in the rope that would prevent the disc from rotating.

So the book's step-by-step answer starts with torque = fd, but they substitute in 30N x 0.25m = 7.5Nm for the torque answer.
However, the equation for torque is one of the forces x perpendicular distance between them.
So shouldn't it be 30N x 0.5m = 15Nm (because the rod is 50cm long)?

Thanks for any replies
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Stonebridge
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(Original post by Rhys_M)
Here is a question from the Oxford OCR student book about couples and torques:

Figure 4 shows a rod of length 50cm with a disc of radius 10cm fixed at its centre. Two forces, both 30N, are applied normal to the rod at each end. Calculate the torque produced by the pair of 30N forces, and then the minimum tension in the rope that would prevent the disc from rotating.

So the book's step-by-step answer starts with torque = fd, but they substitute in 30N x 0.25m = 7.5Nm for the torque answer.
However, the equation for torque is one of the forces x perpendicular distance between them.
So shouldn't it be 30N x 0.5m = 15Nm (because the rod is 50cm long)?

Thanks for any replies
It certainly should be as you say, but is there anything else going on here in the diagram referred to that might have a bearing this? What is that 'disc'? And what about the 'rope'? It would help to see the complete question.
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Rhys_M
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(Original post by Stonebridge)
It certainly should be as you say, but is there anything else going on here in the diagram referred to that might have a bearing this? What is that 'disc'? And what about the 'rope'? It would help to see the complete question.
Name:  90927DB7-B2DB-4573-8F62-EBAB9AC78FDD.jpeg
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Size:  115.4 KBHere is the full question - the highlighted bit.
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Stonebridge
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Rhys_M
Most odd.
The teaching notes clearly state the correct formula, Fd, where d is the perpendicular distance between the parallel forces, and then go on to incorrectly use that in the calculation.
It should indeed be 30 x 0.5 Nm (or 30x0.25 + 30x0.25Nm)
There is no doubt about it.
In the second part, they correctly state that the torque produced by the 2 forces must be balanced by the torque produced by the rope.
And the torque from the rope is correctly stated as T x 0.1 Nm
0.1m is used because the 'couple' in this case (the couple from the rope which opposes the other couple) is created by the force from the rope, and an equal but opposite force of T acting at the pivot. The distance between these 2 is 10cm. There must be a second force acting with the force T from the rope to create a couple.
However, the teaching note seems to have missed this point altogether.
I'm not very impressed with this learning material.
Repost if you need further help with this.
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Rhys_M
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Thank you so much for your help
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