Contact Forces HW Problem with Three Boxes

  • Thread starter physics114
  • Start date
  • #1
Correct? Please let me know if my logic is off. =]

Given Problem
Contact Force with Boxes.jpg

As shown in the figure, a force of magnitude 7.50N pushes three boxes with masses
m1 = 1.30 kg
m2 = 3.20 kg
m3 = 4.30 kg
(a) Find the magnitude of the contact force between boxes 1 and 2. (?N)
(b) Find the magnitude of the contact force between boxes 2 and 3. (?N)

Relevant Equations
F=ma
Strategy?
- Since boxes are in contact, all boxes will have the same acceleration.
- First, find the acceleration given by Newton's 2nd Law. Divide the total horizontal force by the total mass of all three boxes.
- Secondly for (a) and (b) Find individual contact forces by multiplying individual mass with the acceleration found previously.

Attempted Solution
Find contact acceleration
F = ma
a = F/m
a = (7.50N) / (m1+m2+m3)
a = (7.50N) / (1.30kg + 3.20kg + 4.30kg)
a = (7.50N) / (8.8kg)
a = .852273 m/s^2

Part (a)
force between box 1 and box 2
F = ma
F = (3.20kg)*(.852273m/s^2)
F = 2.7272N

Part (b)
force between box 2 and box 3
F = ma
F = (4.30kg)*(.852273m/s^2)
F = 3.66477N

For both parts a and b I was following the example in the book, but don't really understand why we would be multiplying the acceleration with the mass of the box to the right instead of the left. Is this because the force is being applied from left to right and in order to find the contact force (pushing against the overall force) we use the mass of the box exerting the contact force??
 

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,939
1,390
Correct? Please let me know if my logic is off. =]

Given Problem
View attachment 29202
As shown in the figure, a force of magnitude 7.50N pushes three boxes with masses
m1 = 1.30 kg
m2 = 3.20 kg
m3 = 4.30 kg
(a) Find the magnitude of the contact force between boxes 1 and 2. (?N)
(b) Find the magnitude of the contact force between boxes 2 and 3. (?N)

Relevant Equations
F=ma
Strategy?
- Since boxes are in contact, all boxes will have the same acceleration.
- First, find the acceleration given by Newton's 2nd Law. Divide the total horizontal force by the total mass of all three boxes.
- Secondly for (a) and (b) Find individual contact forces by multiplying individual mass with the acceleration found previously.

Attempted Solution
Find contact acceleration
F = ma
a = F/m
a = (7.50N) / (m1+m2+m3)
a = (7.50N) / (1.30kg + 3.20kg + 4.30kg)
a = (7.50N) / (8.8kg)
a = .852273 m/s^2
'Looks good so far. :approve:
Part (a)
force between box 1 and box 2
F = ma
F = (3.20kg)*(.852273m/s^2)
F = 2.7272N
There's a mistake in the above, part (a). See below for a hint on how to correct it.
Part (b)
force between box 2 and box 3
F = ma
F = (4.30kg)*(.852273m/s^2)
F = 3.66477N
Your part (b) is correct! :approve:
For both parts a and b I was following the example in the book, but don't really understand why we would be multiplying the acceleration with the mass of the box to the right instead of the left. Is this because the force is being applied from left to right and in order to find the contact force (pushing against the overall force) we use the mass of the box exerting the contact force??
There's a couple of ways to go about this, but both have something in common:
o you know the object's (or group of objects') acceleration already.
o you know the object's (or group of objects') mass already.
o you now have enough information to solve for the net force acting on that object (or group of objects).

Let's take block m3. Its mass is given in the problem statement. You've already calculated its acceleration. So apply Newton's second law (F = ma) to block m3 in isolation. You know its mass is 1.3 kg and its acceleration is 0.852273 m/s2, so the net force acting on it must be 3.66477 N. If the net force was anything different, block m3's acceleration would be something different. And since there is only one force acting on block m3 (the force coming from block m2), that must be the force between block m2 and m3.

When measuring the force between block m1 and m2 there are a couple of ways to approach the problem. One way is to temporarily treat block m2 and m3 as one big block. Now just repeat what you did above.

Another option is to take block m2 in isolation. You know its mass (which is given in the problem statement). But now there are two forces acting upon it. There is the force coming from block m3 (which you just calculated) and a different force coming from block m1 (which you trying to solve for). Now here is the important part: You know that Newton's second law is ma = ∑F, where there are two forces to be combined on the right side of the equation. You already know block m2's mass and acceleration. Solve for the remaining force (i.e. solve for the force coming from block m1 -- you've already calculated the force coming from block m3 -- and make sure to treat the forces as vectors).
 
  • Like
Likes Brendan Webb
  • #3
60
0
Since all move together a = F/(m1+m2+m3)

Consider the free body diagram of each block

block 1: equilibrium

F-------> m1 <------ X (unknown force contact between 1 and 2)
<---m1a

sum of forces F- X - m1 * a = 0

So X = F - m1a = (m2+m3) * a

block 3: equilibrium:

Y ----> m3
<---- m3a

So contact force between 2 and 3 is Y = m3 * a
 
  • #4
Thank you very much!

Made correction to Part B and treated m2 and m3 as a single block:
F = (m2 + m3) * (a)
F = (3.20kg + 4.30kg) * (.852273m/s2)
F = (7.5kg) * (.852273m/s2)
F = 6.39205 N
 

Related Threads on Contact Forces HW Problem with Three Boxes

  • Last Post
Replies
4
Views
3K
Replies
3
Views
6K
  • Last Post
Replies
4
Views
16K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
1
Views
693
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
7K
Top