i need this explained step by step, im not concerned about the answer,
a uniform rectangular sign is suspended horizontally from two cables, one attached at the left end, the other attached .250 from the right end. the mass of the sign is 3.00 kg and its length is 1.00 m. find the tension in each cable.
2006-11-17 08:01:45 · 3 answers · asked by Amanda R 2 in Science & Mathematics ➔ Physics
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?
Since the sign is stationary, vector sum of forces must be zero. Sketch a free body diagram and label known forces.
2006-11-17 08:06:43 · answer #1 · answered by modulo_function 7 · 1⤊ 0⤋
If the cables are vertical, the problem is no different from a situation where it is resting on a support at one end and another support .25 from the other. Need to pick a pivot point, I choose the support at the end, and sum the CW and CCW torques (negative sign one of them) and set the sum equal to zero. The CW torque would be from the CofM .5 from the pivot, and the CCW torque would be from the support .75 from the pivot.
The above will give you the upward force at the point .75 from the pivot which is the tension on the cable .25 from the other end. That tension will be some fraction of the total weight. The other cable supports the remainder of the sign's weight.
2006-11-17 16:33:58 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
You haven't mentioned what angle the cables are at. If it's not specified then why not assume that the cables are weightless and vertical all you need do then is determine which two tensions sum to the sign's weight (3kg * g) and put a net torque of 0 about the sign's centre of mass.
2006-11-17 16:14:35 · answer #3 · answered by Anonymous · 1⤊ 0⤋