What is the Contact Force between Boxes in a Horizontal Force System?

In summary, three boxes of varying masses are placed side-by-side on a smooth horizontal floor. A 24.0 N horizontal force is applied to the 1.45 kg box, causing it to push against the other two boxes. Using the formula F=ma, the acceleration of the combined system is calculated to be 2.47 m/s^2. By drawing a Free Body Diagram and applying Newton's Second Law, the contact force between the 3.10 kg and 5.20 kg boxes is determined to be 4.06 N.
  • #1
Student3.41
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0

Homework Statement



Three boxes rest side-by-side on a smooth horizontal floor. Their masses are 1.45 kg, 3.10 kg, and 5.20 kg, with the 3.10 kg one in the center. A horizontal force of 24.0 N pushes on the 1.45 kg mass which pushes against the other two masses. What is the contact force between the 3.10 kg and 5.20 kg boxes?


Homework Equations



F=ma

The Attempt at a Solution

 
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  • #2
Student3.41 said:

Homework Statement



Three boxes rest side-by-side on a smooth horizontal floor. Their masses are 1.45 kg, 3.10 kg, and 5.20 kg, with the 3.10 kg one in the center. A horizontal force of 24.0 N pushes on the 1.45 kg mass which pushes against the other two masses. What is the contact force between the 3.10 kg and 5.20 kg boxes?


Homework Equations



F=ma

The Attempt at a Solution

It's F_net = ma. Please show an attempt to find the acceleration of the 3 blocks as they move together. Then go from there.

Welcome to PF!
 
  • #3
PhanthomJay said:
It's F_net = ma. Please show an attempt to find the acceleration of the 3 blocks as they move together. Then go from there.

Welcome to PF!

Ok, well I calculated the Total Mass = 9.75kg, then used the formula F=ma -> 24.0N/9.75kg=2.47m/s^2. I got the acceleration, i then used the same formula to find the original force on M1 = 3.57N, then M1 on M2 = 7.63N which I assumed added together F_2-3=7.63+ -3.57 = 4.06N
 
  • #4
Student3.41 said:
Ok, well I calculated the Total Mass = 9.75kg, then used the formula F=ma -> 24.0N/9.75kg=2.47m/s^2. I got the acceleration,
this is correct
i then used the same formula to find the original force on M1 = 3.57N,
that's the NET force on M1
then M1 on M2 = 7.63N which I assumed added together F_2-3=7.63+ -3.57 = 4.06N
You are not drawing Free Body Diagrams (FBD's). Look at block 3, the 5.2 kg block. Isolate it from the rest of the blocks and determine the forces acting on it in the x direction. There is only one force acting on it, the contact force from the middle block accelerating it forward. Calculate that force using Newton 2.
 
  • #5
Thank You, I ended up figuring it out with a FBD.
 

1. What is a contact force?

A contact force is a force that results from the physical contact between two objects. It can either be a push or a pull and is exerted by one object on another.

2. What are some examples of contact forces?

Some examples of contact forces include friction, normal force, tension, and air resistance.

3. How do contact forces affect the motion of objects?

Contact forces can cause objects to accelerate, decelerate, or change direction depending on the direction and magnitude of the force. They can also cause objects to deform or break if the force is strong enough.

4. How do you calculate the magnitude of a contact force?

The magnitude of a contact force can be calculated using Newton's Second Law of Motion, which states that force is equal to mass multiplied by acceleration. So, the magnitude of a contact force is directly proportional to the mass of an object and the acceleration it experiences.

5. What factors can affect the strength of contact forces between boxes?

The strength of contact forces between boxes can be affected by the mass, surface area, and surface texture of the boxes. The type of material the boxes are made of and the presence of any lubricants or adhesives between the surfaces can also impact the strength of contact forces.

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