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Grade 11Mechanics

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, m₁ of mass 1.2 kg and m₂ of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m₂ is in contact with the floor. (a) What is the largest value the force may have so that m₂ 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 m₁?

Simran Bhatia

11 Years agoGrade 11

2 Answers

Aditi Chauhan

11 Years ago

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 .
(b)
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.

Ashutosh Garg

11 Years ago

a. Taking the same situation mid air we get that m₂ goes down with the accelaration of a = (m₂-m_1)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

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