particles of mass 3m and 5 m hang one at each end of a light inextensible string which passes over a pulley.The system is released from rest with the hanging parts taunt and vertical.During the subsequent motion the resultant force exerted by the string on the pulley is of magnitude-
a 8mg
b 7.5mg
c 3.75mg
d 4mg
e 2mg
Machanics? taunt?
Wow.
Now to greater errors. What exactly is a mass of 3m or 5m?
And in the answers, since when is force measured in milligrams? Isn't that a mass unit?
Something is major wrong in your physics class, this usage is not acceptable: Remember this was the problem that brought down the English units, confusing lbs force and lbs mass.
I am not taunting you or your instructor, I hope.
To find the magnitude of the resultant force exerted by the string on the pulley, we need to consider the forces acting on each mass individually.
Let's label the mass of the particle at one end of the string as m1 = 3m, and the mass of the particle at the other end as m2 = 5m.
When the system is released, the heavier mass (m2 = 5m) will start to descend due to its weight, while the lighter mass (m1 = 3m) will start to ascend. At the same time, the string passes over a pulley, which will accelerate due to the difference in weights.
For the mass m1, the direction of the force exerted by the string is upwards (against its weight). So, the force exerted by the string on m1 is equal to the weight of m1, which is F1 = m1 * g.
For the mass m2, the direction of the force exerted by the string is downwards (in the direction of its weight). So, the force exerted by the string on m2 is equal to the weight of m2, which is F2 = m2 * g.
Now, let's consider the pulley. Since the string passes over the pulley, the direction of the force exerted by the string on the pulley will depend on the relative magnitudes of F1 and F2.
If F1 is greater than F2, the pulley will rotate in the clockwise direction. If F2 is greater than F1, the pulley will rotate in the counterclockwise direction.
In this case, since m2 is greater than m1 (5m > 3m), F2 will be greater than F1. Therefore, the pulley will rotate in the counterclockwise direction.
The resultant force exerted by the string on the pulley is equal to the difference between F2 and F1, which is F_res = F2 - F1.
Substituting the values, we have:
F_res = (m2 * g) - (m1 * g)
F_res = (5m * g) - (3m * g)
F_res = 5mg - 3mg
F_res = 2mg
Therefore, the magnitude of the resultant force exerted by the string on the pulley is 2mg.
Hence, the correct option is e) 2mg.