• Support PF! Buy your school textbooks, materials and every day products Here!

Newton's 3rd Law - block on top of another block

  • Thread starter banana2020
  • Start date
  • #1

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 cancelled 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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,146
491

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 cancelled 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?
 

Related Threads on Newton's 3rd Law - block on top of another block

Replies
7
Views
1K
Replies
3
Views
6K
Replies
5
Views
5K
  • Last Post
Replies
19
Views
6K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
0
Views
834
  • Last Post
Replies
1
Views
962
  • Last Post
Replies
8
Views
482
Replies
3
Views
3K
Replies
1
Views
2K
Top