Question

Problem 7:
The two masses 1 and 2 are connected to the mass 3 by a cord passing over a pulley having
a negligible mass. The coefficient of friction of the inclined surface is = 0.12 and 1= 5 Kg,
2= 2 Kg .
1. What should be the mass 3 so that the two masses on the incline will move down with
an acceleration of 2 m/s².
2. In this configuration, what are the Tensions 1 and 2 ?

Answer 1

Answer:

system will accelerate at same acceleration.

a ( m1+m2+m3) = m3g - FF

FF = ( m1+m2) g . 0.12 . cos Theta.

Theta is Missing in problem data. a = 2 .

plug value of Theta above to Get FF and hence M3

depending on position of m1 is ahead or m2 is ahead

m1 . a = T1 - mg Sin Theta.

get T1 above.

m3.a = T2 - M3 g

get T2 above.

( i have modeled this for All problems of this type . thanks to Missing picture.)