Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
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, 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?
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
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details