When two objects of unequal mass are hung vertically over a frictionless pulley

Question

When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in the figure, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to calculate the value of g. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord.

Suppose that in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 6.15 kg and m2 = 10.85 kg, what value of f will reduce the acceleration of the system by 53%? f = N

Explanation / Answer

ind the acceleration of the system without the force F: up is the positive direction, so we have for forces on the masses: T - m1 g = m1 a T - m2 g = - m2 a subtract equations to get: (m2-m1)g = (m1+m2)a => a= (m2-m1)g/(m2+m1) = g/3 we need a force that will cause the acceleration of the system to be g/6 the newton's second law equations become: T - m1 g - f = m1a T- m2 g = - m2 a subtract equations: (m2-m1)g -f = (m1+m2)a f= (m2-m1)g - (m1+m2)a remembering that a=g/6, use the masses and solve for f

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.