Question

Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M (μs = 0.26, μk = 0.15) so the three blocks accelerated together to the right.


What is the acceleration of the blocks? (You may assume block 3M does not slide or fall off block 2M.)

What is the maximum force F that can be applied, before the 3M block slides off?

Answer 1

Part 1: If block 3M does not slide or fall off block 2M when a constant force of 28 N is applied to block M then by Newton's second law
\[ F = (M+2M+3M)a = 6Ma\]
implying that the acceleration on each block is of the amount
\[ a = F/(6M)\]

Answer 2

Thanks!

Answer 3

Part 2: For sliding to occur, the applied force (F) to block M must be greater than the frictional force (Ff) between block 2M and 3M. That is,
\[ F > F _{f} = \mu _{t}N \] where \[ \mu _{t} = \mu _{2M} +\mu _{3M} = 0.26+0.15 \] The normal force (N) is equal to the weigth of block 3M by the relation \[ N= F _{3M} =3Mg\] (check out the drawing).
Therefore \[ F > F _{f} = \mu _{t}N =3\mu _{t}Mg\] where g is the gravitational acceleration of the Earth. So to summarise, for block 3M to begin sliding off block 2M the applied force F must be greater than \[ F > 3\mu _{t}Mg\]
dw:1329607904844:dw

Answer 4

Thanks again!

Answer 5

no probs :)