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Grade: 11


Someone exerts a force F directly up on the axle of the pulley shown in Fig. Consider the pulley and string to be mass- less and the bearing frictionless. Two objects, m 1 of mass 1.2 kg and m 2 of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m 2 is in contact with the floor. (a) What is the largest value the force may have so that m 2 will remain at rest on the floor? (b) What is the tension in the string if the upward force F is 110 N? (c) With the tension determined in part (b), what is the acceleration of m 1 ?

5 years ago

Answers : (2)

Aditi Chauhan
askIITians Faculty
396 Points
Round off to two significant figures,
F = 37 N
Therefore, the maximum force F for which block with mass m2 stays on floor is 37 N .
The tension in the string is related to upward force as:

Therefore, the tension in the string corresponding to upward force of 110 N is 55 N.

5 years ago
Ashutosh Garg
25 Points
a. Taking the same situation mid air we get that m2 goes down with the accelaration of a = (m2-m1)g/(m2+m1).
     Now considering this case for the mass m2 to remain at the ground the pulley shold go up 2a.
     so F=(m1+m2)2a = 2(m2-m1)g. putting values we get the answer = 14N
pls aprove if it is helpful to you
5 years ago
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