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

AI Thread Summary
In a horizontal force system with three boxes of masses 1.45 kg, 3.10 kg, and 5.20 kg, a horizontal force of 24.0 N is applied to the 1.45 kg box. The total mass of the system is calculated to be 9.75 kg, resulting in an acceleration of 2.47 m/s². The contact force between the 3.10 kg and 5.20 kg boxes is determined by analyzing the forces acting on the 5.20 kg box using Free Body Diagrams (FBDs). The calculations show that the contact force is derived from the net forces acting on the boxes as they accelerate together. Ultimately, the problem is solved effectively by applying Newton's second law and utilizing FBDs.
Student3.41
Messages
70
Reaction score
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

 
Physics news on Phys.org
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!
 
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
 
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.
 
Thank You, I ended up figuring it out with a FBD.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Contact Force with two masses given
Replies
4
Views
4K
How Does Pushing Force Affect Contact Force Between Two Boxes?
Replies
9
Views
11K
Newton's law problem: Pushing 2 stacked blocks on a horizontal table
Replies
13
Views
3K
Contact force of block on a wall
Replies
5
Views
3K
Tension in a string between two blocks
Replies
45
Views
23K
Calculating Contact Forces- Three objects
Replies
3
Views
8K
Basic Newton's Laws and Applied Force: 3 boxes
Replies
5
Views
8K
Distribution of Energy when work is done on a system of 2 masses connected by a spring
Replies
17
Views
1K
A question about contact forces and friction
Replies
6
Views
5K
Calculating Force to Push Steel Rod Up 50 cm
Replies
24
Views
3K

Hot Threads

Recent Insights

Back
Top