# Physics: Tension on a string from a system [closed]

Ok, I am stumped by this question:

In the system in the diagram below, block M (15.7 kg) is initially moving to the left. A force F , with magnitude 60.5 N, acts on it directed at an angle of 35.0 degrees above the horizontal as shown. The mass m is 8.2 kg. There is no friction and the pulley and string are massless.

1. What is the normal force (N) on M ?
2. What is the tension (N) in the string?
3. What is the acceleration (m/s2) of M (positive is to the right)?

I tried and got the following:

Got the X component of the force F: F * Cos (35) = 49.55869868 N
Got the Y component of the force F: F * Sin (35) = 34.7013744 N
Got the force of m on M: 8.2 * 9.8 = 80.36 N (which is what I assumed the tension on the string was)

I got the acceleration of M caused by m = -5.118471338
Then I got the acceleration of M cause by F = 3.156605011

Putting it all together I got the following answers:

1. 129 N
2. 80.4 N
3. -1.96 m/s^2

I (obviously) got all three wrong. The correct answers are:

1. 119 N
2. 69.8 N
3. -1.29 m/s^2

So, I'm stuck. There is something missing that is causing my string tension to be off. If I would have gotten the right string tension, all the other answers would have been correct. I know that is where the problem is, so if someone could point out my mistake, that would be wonderful!

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## closed as too localized by dmckee♦May 1 '11 at 14:04

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Welcome to phyisc.SE. We don't "solve homework problems" here, though we will answer questions about the physics that are bothering you. This, of course, means that you have to figure out what physics is troubling you, which may be equivalent to solving the problem. – dmckee May 1 '11 at 14:03
May I suggest that you look again at the tension in the string. IN particular look at the free body diagram only of the bob, noting two forces weight $mg$ down and Tension $T$ up, and that the bob experiences a non-zero vertical acceleration. – dmckee May 1 '11 at 14:05

For 1, you should have an normal force of $Mg −F \sin \theta$ which being positive means that the large mass does not move vertically.

For 3, you know the net accelerating force (upwards for the small mass or to the right for the large mass) on the two masses combined is $F \cos \theta - mg$, and you know the combined mass $m+M$ so you can calculate the acceleration $a$ in the same directions.

For 2, your answer to 3 has shown that the small mass is accelerating downwards by less than $g$ (or upwards by more than $-g$), so there is an upward force on the small mass caused by tension in the rope, and this is (watch out for signs) $m(a-(-g))$.

If you get the arithmetic correct this should give you the right answers.

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If m is accelerating, the tension in the string is not $mg$, but $m(g-a)$ where $a$ is the acceleration. If there were no string, there would be no tension, and the mass would be accelerating at $g=9.8 m/sec^2$. So you need to calculate the total acceleration on M, which will also be the acceleration on m, then reduce the string tension appropriately.

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