Solving for Acceleration in an Incline Plane Problem: m1=4.50kg, m2=2.80kg

Click For Summary
SUMMARY

The problem involves two blocks, m1 = 4.50 kg on a frictionless incline at 30.0° and m2 = 2.80 kg hanging vertically, connected by a cord over a massless, frictionless pulley. The acceleration of both blocks is the same in magnitude but opposite in direction. The correct approach requires applying Newton's second law, where the tension T in the rope must be calculated based on the forces acting on each mass. The relationship T - m1*g = m1*a is crucial for determining the acceleration.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of free body diagrams
  • Familiarity with inclined plane physics
  • Basic algebra for solving equations
NEXT STEPS
  • Study the principles of Newton's second law in detail
  • Learn how to draw and analyze free body diagrams
  • Explore the dynamics of frictionless inclined planes
  • Practice solving problems involving pulleys and connected masses
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for problem-solving strategies in inclined plane scenarios.

maff is tuff
Messages
65
Reaction score
1

Homework Statement



A block of mass m1 = 4.50 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.80 kg hanging vertically.

What is the magnitude of the acceleration of each block?



Homework Equations





The Attempt at a Solution



I have attached two of my many attempts. I think the second attempt is closer because I realized that since the blocks are connected that they have the same magnitude of acceleration but just different directions. Both of these answers are wrong along with the many others I have submitted. Some help and/or hints would be great. Thanks.
 

Attachments

Physics news on Phys.org
The force acting on the mass m1 via mass m2 is the tension T in the rope and it is only equal to m_1*g when nothing moves. The mass m_1 is accelerating so T - m_1*g = m_1*a, T is not equal to m_1.

I think it helps to draw a separate free body diagram for each mass.
 

Attachments

  • mass on plane027.jpg
    mass on plane027.jpg
    18.2 KB · Views: 469
Ok thanks that makes sense. I'll try the problem again tomorrow.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
9
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K