- #36
- 22,183
- 3,325
Headacheguy said:A 2 kg body is supported by a massless string. The body is accelerating downwards at a rate of 4.9 m/s^2. Calculate the tension on the string.
(It was pretty odd that this question is in the midterm exam and is worth 10 points.)
What I did was:
From F = ma:
T + (-mg) = -ma (Took the tension T as positive, therefore the weight -mg, is negative since the force is exerted on the opposite direction. The resultant force of this body is downwards, same direction with the weight, as stated in the question, -ma.)
T = mg - ma
T = 2 kg ( 9.8 m/s^2 - 4.9 m/s^2)
T = 9.8 N
T = 10 N
Her solution was a bit different because she chose the direction of the motion of the body as the positive direction. Either way, the answer should have been the same. Hence, the weight is positive, so is the direction of the resultant force, and the tension negative because it is on the upwards direction.
mg - T = ma
T = mg - ma
T = m (g - a) (This was where it went wrong, I think)
T = 2 kg [ 9.8 m/s^2 - (-4.9 m/s^2)] (The acceleration was negative because it was going "down", she explained)
T = 29.4 N (Again her answer was not in the correct significant figures.)
Most of my classmates had the same answer as mine.
Please forgive me if my solution is wrong.
Hahaha! Ask her: what if the body was accelerating downwards with a rate of more than g. Intuitively, it should put no tension on the string, and your answer indeed gives that. Put her answer would be that it put more tension on the string than without acceleration!