Question

in the given figure the pulley is assumed mass less and friction less if the friction force on the object of mass M is f then its acceleration in terms of the force F will be equal to

Answer 1

Answer:

Free body diagram of m

W + T '3 = 0

W = (0 , -|W|) , weight

T '3 = (0 , |T '3|) , tension of string

y components equation gives

|T '3| = |W|

Tension in the same cord

|T3| = |T '3| = |W|

2) Free body diagram of point P

At point p (origin of x y system of axes)

T1 + T2 + T3 = 0 (Newton's second law)

T1 = ( |T1| cos α1 , |T1| sin α1)

T2 = ( -|T2| cos α2 , |T2| sin α2)

T3 = (0 , -|T3|) = (0 , -|W|) , (since |T3| = |T '3| = |W| already found above)

sum of x components = 0 gives equation

|T1| cos α1 - |T2| cos α2 = 0 (equation 1)

sum of y components = 0 gives equation

|T1| sin α1 + |T2| sin α2 - |W| = 0 (equation 2)

The above is a system of 2 equations with 2 unknown |T1| and |T2|.

Solving the system of equations using Cramer's rules, we obtain

|T1| = |W| cos α2 / ( cos α1 sin α2 + sin α1 cos α2 ) = |W| cos α2 / sin(α1 + α2)

|T2| = |W| cos α1 / ( cos α1 sin α2 + sin α1 cos α2 ) = |W| cos α1 / sin(α1 + α2)

Answer 2

$\frac{f}{2} - f$

right answer...