Finding tension when upper block slides under lower block

[banana2020]

Homework Statement

A m1=5.35 kg block is placed on top of a m2=11.6 kg block. A horizontal force of 89.7 N is applied to the 11.6 kg block causing it to slide. The 5.35 kg block is tied to the wall by a string. The co-efficient of kinetic friction between all surfaces is 0.247. Determine the magnitude of the tension in the string. Find the acceleration of the 11.6 kg.

Homework Equations

F = ma

The Attempt at a Solution

In order to find tension in the string, i attempted to find the net acceleration and below is my approach...

I drew a FBD separately for m1 and m2.
The net acceleration is to the right for m2.

FBD for m2 (lower):

the applied force is in the right direction = 89.7 N
the opposing force (Fk) is = µmg = µ(m1 + m2)g = µ(16.95)(9.81) = 48.67 N
F - Fk = ma
a = (89.7 - 48.67)/16.95 = 2.87 m/s²

my thinking is that the applied force is applied to the lower block, so it must also include the mass of the upper block. thus the force of kinetic friction (in the opposite direction) must also include the mass of the upper block and the lower block...

i tried to find the vertical component of the acceleration but the forces canceled out and i ended up with a in y direction = g


FBD for m1:
since the net acceleration of m1 is to the left, the T and Fk are to the left side.
T + Fk = ma
T = ma - Fk = (5.35)(2.87) - µ(5.35)(9.8) = 2.41 N

 

[PhanthomJay]

Homework Statement

A m1=5.35 kg block is placed on top of a m2=11.6 kg block. A horizontal force of 89.7 N is applied to the 11.6 kg block causing it to slide. The 5.35 kg block is tied to the wall by a string. The co-efficient of kinetic friction between all surfaces is 0.247. Determine the magnitude of the tension in the string. Find the acceleration of the 11.6 kg.

Homework Equations

F = ma

The Attempt at a Solution

In order to find tension in the string, i attempted to find the net acceleration and below is my approach...

I drew a FBD separately for m1 and m2.

Good!

The net acceleration is to the right for m2.

Yes!

FBD for m2 (lower):

the applied force is in the right direction = 89.7 N
the opposing force (Fk) is = µmg = µ(m1 + m2)g = µ(16.95)(9.81) = 48.67 N
F - Fk = ma
a = (89.7 - 48.67)/16.95 = 2.87 m/s²

, No, there is also an opposing friction force between the 2 blocks.

my thinking is that the applied force is applied to the lower block, so it must also include the mass of the upper block.

In your FBD, the Normal force (not the mass) of the top block must be included.

... thus the force of kinetic friction (in the opposite direction) must also include the mass of the upper block and the lower block...

There are 2 kinetic friction forces acting on the lower block, each of which is a function of the appropriate normal force, not the mass.

i tried to find the vertical component of the acceleration but the forces canceled out and i ended up with a in y direction = g

but there is no a in the y direction...Newton 1.

FBD for m1:
since the net acceleration of m1 is to the left, the T and Fk are to the left side.
T + Fk = ma
T = ma - Fk = (5.35)(2.87) - µ(5.35)(9.8) = 2.41 N

Homework Statement

The top block does not move..it thus has no acceleration with respect to the ground (Newton 1 again). Per Newton 3, what is the direction of the friction force on the top block?