Question

Two masses, m_1m ​1 ​​ = 3.75 kg and m_2m ​2 ​​ = 6.34 kg, are connect by a string of negligible mass. The string passes over a frictionless pulley so that m_1m ​1 ​​ and m_2m ​2 ​​ hang down on opposite sides of the pulley. The whole system is released from rest. Calculate the tension in the string as the masses are moving.

Answer 1

Step-by-step explanation:

For the mass m1;

The sum of forces acting on the body is expressed according to Newton's second law as;

\sum Fx = m1ax

T - Ff = m1a .... 1

T is the tension

Ff is the frictional force acting on m1

For the mass m2:

\sumFy = m2a

W - T = m2a

m2g - T = m2a.... 2

W is the weight

g is the acceleration due to gravity

If the acceleration of the system is 0, the equation becomes;

From 2:

m2g - T = m2a.... 2

a = m2g-T/m2

From 1:

a = T-Ff/m1

Equate both accelerations

m2g-T/m2 = T-Ff/m1

Cross multiply

m1m2g - Tm1 = m2T-m2Ff

m1m2g+m2Ff = m2T+m1T

m1m2g+m2Ff = T(m2+m1)

T = m1m2g+m2Ff/m2+m1

Hence the tension in strings is expressed as

T = m1m2g+m2Ff/m2+m1