# Coupled Motion Pulley Problem

1. Sep 30, 2007

### johnsonandrew

This has been frustrating me:

Problem:

The pulley in the figure is essentially weightless and frictionless. Suppose someone holds on to m=10.0 kg and accelerates it upward at 4.905 m/s/s. What will then be the tension in the rope given that M=10.0 kg?

Attempt:

The sum of the forces on m = Ftension + Fhand - Fgravity = ma
The sum of the forces on M = Ftension - Fgravity = Ma

I said Fhand= m * upward acceleration

I tried solving for 'a' on both equations, then set them equal to eachother to solve for Ftension. This gave me a ridiculous answer, and I'm pretty sure I didn't screw up my math. I must have set it up wrong? Please help!

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Last edited: Sep 30, 2007
2. Sep 30, 2007

### AbedeuS

I would guess you calculate the tension in the string for when the Weight is completely motionless, this is equal to:

$$F=mg = 10*9.8 = 98N$$

With this tension in mind, this will always be persistant through your calculations, now the mass is being accelerated upwards at a acceleration of 4.905m/s^2 so:

$$F=ma = 4.905*10 = 49.05N$$

Total reactive tensile force in the rope is equal to the force of gravity pulling downwards which causes a tension upwards and opposite to it plus the force applied to the weight to accelerate it upwards these tension forces are in the same direction, imagine, the tension force that is reactive to gravity, being larger then the force of gravity, things on pullies would magically float upwards, so it follows that the reactive tension force is equal and opposite to gravity, and the upwards accelerative force is in the same direction of the tension force that prevents gravity otherwise this accelerative force would be pushing the weight downwards, which is counterintuitive:

$$F_{total} = F_{gravity}+F_{acceleration}=49.05+98=147.05N$$

Last edited: Sep 30, 2007
3. Sep 30, 2007

### johnsonandrew

The second answer, 49.1 N, is the answer the book gives. But I'm confused as to why you calculate the tension when the weight is motionless first. And aren't you neglecting the force the other mass, M, in calculating the tensile force? The answer 147.05 N is incorrect, says the book.

4. Sep 30, 2007

### johnsonandrew

Ohhh you know what? I think the acceleration given becomes the acceleration of the system. So a = 4.905. Why didn't I think of that before! I'm making it more complicated than it is, making two seperate accelerations for some reason. That is the problem isn't it?

5. Sep 30, 2007

### johnsonandrew

But then I get 147 N... Damn .. What am I doing wrong..