Spring Balance readings

Question

a pulley of negligible weight is suspended by a spring balance '8'. Masses of 3 kg and 7 kg respectively are attached to opposite ends of a string passing over a pulley 'P'. The spring balance reads'?

Answer 1

Hello Prashant,

I'm afraid that Steven W. made a mistake. If you set T equal to 51.16 N in the first equation, T - 29.4 N = 3a, and solve for 'a', you get  a  =  7.253 m/s^2. If you set T equal to 51.16 N in the second equation, T - 68.6 N  =  -7a, and solve for 'a', you get  a  =  -2.49 m/s^s. You would expect one acceleration to be positive and the other to be negative, because one mass is accelerating upward and the other is accelerating downward, but their magnitudes should be the same, not different. Because the two masses are attached to each other by the string and must move with the same acceleration (sign ignored), you should treat the two masses as a single mass of 10kg. The net force acting on this composite object is  68.6 N - 29.4 N  =  39.2 N, when looking from the point of view of the 3kg mass, and 

29.4 N - 68.6 N  =  -39.2 N, when viewing the system from the 7kg mass's viewpoint. From the equation for acceleration, a  =  F_net / m, we get  a  =  39.2 N / 10 kg  =  3.92 m/s^2 for the 3kg part of the composite mass, and  a  =  -39.2 N / 10 kg  =  -3.92 m/s^2  for the 7kg part of the composite mass. Now our accelerations have the same magnitude, and opposite signs. Using the same equation, rearranged, F_net   =  ma, for each of the masses, and the equation,  F_net  =  F_T  -  F_G, we find that the force of tension on the string, F_T, is equal to 41.16 N.

Now, as Steven W said, this force of tension is being applied to both sides of the pulley, and the pulley is at translation rest (note: not at rotational rest), the force exerted by the spring balance should equal twice the force of tension, or 82.32 N. This is less than what you might expect the spring balance to read, 10 kg * 9.8 m/s^2  =  98. N. Here is my follow-up question to you: why does the spring balance read less than 98. N ?

 

Chris Jones

 

Answer 2

Hi Prashant!

 

If we string the two indicated masses over the pulley as described, they will accelerate such that the 7 kg mass falls and the 3 kg one rises.  This will create a certain tension in the string (which I assume is massless, as these problems generally assume, in the absence of other information).  I would suggest solving for that tension first.  We can use a system of two equations generated by setting up Newton's 2nd law on each mass:

 

[note:  I assume downward is negative for each mass]

 

For the 3 kg mass (accelerating upward):

 

F_net = T - mg = ma -->  T - (3 kg)(9.8 m/s²) = (3 kg)a --> T - 29.4 = 3a

 

For the 7 kg mass (accelerating downward):

 

F_net = T - mg = m(-a) --> T - (7 kg)(9.8 m/s²) = -(7 kg)(a) --> T - 68.6 = -7a

 

If the string is massless, T is the same throughout it, and thus the same tension is applied to both  masses.  And, if the string is taut, they must have the same acceleration, since both must move exactly the same way over time.  This means the a and the T in both of the above equations is the same a and T.  We thus have a system of two equations with two unknowns, which we can solve for T.

 

I will solve that by multiplying the 3-kg equation by 7, the 7-kg equation by 3, and then adding them together.  This gives:

 

7T - 205.8 = 21a

3T - 205.8 = -21a

---------------------

10T - 411.6 = 0

 

10T = 411.6

 

T = 41.16 N

 

Now, this tension is applied downward on the pulley twice, once pulling down on each side.  Whatever is holding the pulley up has to provide a force -- read by the spring balance -- that cancels out that total downward force from the tension (on each side).  So, we can say:

 

F_{net-on-pulley} = F_up - 2T = F_up-82.32 N = 0  (since the pulley is in translational equilibrium)

 

So F_up, which the spring balance reads, will equal 82.32 N (a bit more than if it were just holding up stationary masses).

 

Hope this helps!  I will check my calculations again later, but the path to solution is, I think, sound.  If you have any questions about it, just let me know!