Question

Two masses are suspended by cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m1, has a mass of 7.0 kg and the other mass, m2, has a mass of 3.0 kg. The pulley turns on a shaft through the center of the pulley and supports the pulley and all the masses. The vertical force of the shaft on the pulley that supports the whole system is

Answer 1

Answer: F = 98N

Explanation: The shaft have to sustain the pulley, the cord and the two masses. The pulley and the cord have negligible masses, so, they have negligible weight.

The two masses have two vertical forces acting on them: force of traction because of the cord and force due to gravitational force, also known as weight.

So, the vertical force the shaft has to support is the sum of the weight of each mass:

$F_{net}=F_{g}_{1}+F_{g}_{2}$

$F_{net}=m_{1}.g+m_{2}.g$

$F_{net}=g(m_{1}+m_{2})$

$F_{net}=9.8(7+3)$

$F_{net}=$ 98

The vertical force that supports the whole system is 98 N.