Determining the Acceleration (Two objects in contact)

Click For Summary
SUMMARY

The discussion focuses on calculating the acceleration and forces between two crates in contact, specifically a 75kg crate (crate x) and a 110kg crate (crate z) subjected to a 620N force. The total mass of the system is 185kg, resulting in an acceleration of 3.35 m/s². The force exerted by crate x on crate z is 251.25N, while the force exerted by crate z on crate x is 368.5N. The calculations remain consistent when the crates are reversed, emphasizing the importance of accounting for friction in the analysis.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of friction coefficients
  • Ability to perform vector force analysis
  • Familiarity with free body diagrams
NEXT STEPS
  • Explore advanced applications of Newton's laws in multi-body systems
  • Study the effects of varying friction coefficients on acceleration
  • Learn about force diagrams and their role in physics problem-solving
  • Investigate the impact of mass distribution on system dynamics
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding dynamics and forces in contact systems.

yandereni
Messages
13
Reaction score
1
1. The problem statement, all variables and given/known data
2 crates, 75kg(crate x) and 110kg(crate z) are in contact and at rest on a horizontal surface. A 620N force is applied on crate x . If the coefficient of friction is 0.15, calculate:
a.) the acceleration of the system
b.)the force that each crate exerts on each other
c.) repeat when the crates reversed

2. Homework Equations

Msys = (massx)+(massz)
Force = mass(acceleration)

The Attempt at a Solution



a. Msys = 75kg + 110kg
= 185kg
620N = 185kg(a)
a = 3.35m/s2

b.
cratex= 75kg(3.35m/s2)
=251.25N
cratez= 110(3.35m/s2)
=368.5N

c.
cratez= 110(3.35m/s2)
=368.5N
cratex= 75kg(3.35m/s2)
=251.25Nthanks in advance!
 
Physics news on Phys.org
It may help to draw a force diagram to help yourself with this. For instance, with part A, you have to include friction as a part of what's going on with the system. So when you add up all the forces, you have a slightly different answer than what you have because the frictional force reduces the net force on the blocks in the direction of their displacement, and therefore reduces the acceleration. This is also what you have to keep in mind for parts B and C.
 

Similar threads

Replies
6
Views
25K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
5K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
3
Views
1K
Replies
10
Views
5K
  • · Replies 13 ·
Replies
13
Views
5K