Halliday, Resnick & Krane Chapter 5: Force on Pulley

In summary, the tension on a pulley can be found by adding the equations for m1 and m2, resulting in 2T + N = (a+g)(m1+m2). The masses will have the same acceleration and the next step is to solve for T. A hint for solving for T would be to use the equations for m1 and m2 to eliminate variables.
  • #1
vibha_ganji
19
6
Homework Statement
Someone exerts a force F directly up on the axle of the pulley shown in Fig. 5-45. Consider the pulley and string to be massless 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 FB 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?
Relevant Equations
F=ma
Force on pulley = 2(tension)
As the force on a pulley is equal to twice the tension, I just have to find the tension to solve part A. To do so, I first wrote the equations for both m1 and m2.

m1 * a = T - m1g

m2 * a = T + N - m2g

The tension must have the same values for both equations so I added both equations to find the tension.

m1a + m2a = 2T + N - g(m1 + m2)
a(m1+m2) + g(m1+m2) = 2T + N

(a+g)(m1+m2) - N = 2T

I’m not sure what to do next. Can I have a hint?
F3E23044-B23C-487C-A296-375AD1023D15.jpeg
 
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  • #2
vibha_ganji said:
m1 * a = T - m1g

m2 * a = T + N - m2g
How are you defining a? Will the masses have the same acceleration?
 

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