Torque and Net Force Calculations

  • #1

Homework Statement


The bar, of length AB=2m, with a mass of 6.00 kg.

There are two weights, lets call them m1 and m2, hanging off the bar. m1 is 4.4kg, at 0.66m from point A. m2 is 2.0 kg, at 1.84m.

The bar is suspended by two spring scales defined as F1 and F2 (assuming negligible mass). The first spring scale, F1, is 68.2N from a distance of 0.45m from A. The second spring scale is 56.3N from a distance of 1.71m.

This system is at equilibrium, therefore, net forces and net torque must be zero. I say must be, because the

Homework Equations


ΣFx = ΣFy = 0 T=Fxd = mass x acceleration x d

3. Questions posed;

a) calculate ΣFx
b) calculate ΣFy
c) calculate Torque at A
d) calculate the percentage error for ΣFy and Torque at A (a very strange question for what is essentially a theoretical math question)

4. The attempt at a solution

Given that there is no reference to horizontal forces in the above description (no wires attached to walls, etc.), we deemed ΣFx as zero.

ΣFy, however, was 0 = 68.2N + 56.3N - 43.1N - 58.8N - 19.6N = 3N....which is a percentage error, but that's a divide by zero calculation.

Torque at A was calculated = 68.2*0.45 + 56.3*1.71 - 43.1*0.66 - 58.8*1.00 - 19.6*1.84 = 3.653Nm....which makes very little sense given that the whole system should be spinning to the left.

upload_2016-1-16_22-40-0.jpeg

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Answers and Replies

  • #2
haruspex
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My guess is that the sums of forces and torques asked for is just those due to the rod and masses. So the percentage error being sought is not the error in the net force, it is the error in the total force measured by the spring scales. Likewise torque.
 

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