Consider two blocks (M 1 and M 2 ) connected by the same rope passing over a mas
ID: 1593778 • Letter: C
Question
Consider two blocks (M1 and M2) connected by the same rope passing over a massless and frictionless pulley, but with one mass resting on a plane inclined 30 degrees to the horizontal (30/60/90 degree triangle).
1. If the masses of the blocks are equal, what is the acceleration of block M2 assuming the interface between mass M2 and the inclined plane to be frictionless? (Be sure to indicate whether it moves up or down the ramp).
2. What must be the mass of M2 be in relation to mass M1, in order to move in the opposite direction along the ramp?
3. If mass M2 were one-tenth the value of mass M1, what coefficient of friction would be required to hold the blocks in place with no sliding?
4. What is the tension in the string under this condition?
Explanation / Answer
1.
assume that M2 is moving down and acceleration of the system is a m/s^2
let tension in the rope be T
then writing force balance equation for M2:
M2*g*sin(30)-T=M2*a
==>M2*(4.9-a)=T...(1)
writing force balance equation for M1:
T=M1*a...(2)
from equation 1:
M2*4.9-M2*a=M1*a
==>a=4.9*M2/(M1+M2)
Q2. as a=4.9*M2/(M1+M2) unless M2 is negative, acceleration can not be negative and hence can not change direction
Q3. let coefficient of friction be k and the blocks are not moving
then for block M1:
T=friction on block M1=k*M1*g...(3)
for block M2:
M2*g*sin(30)=T+k*M2*g*cos(30)
==>M2*4.9=k*M1*9.8+k*M2*8.487
==>k=4.9*M2/(9.8*M1+8.487*M2)
Q4. tension=k*M1*9.8
=48*M1*M2/(9.8*M1+8.487*M2)