Tension of 2 Bodies on a Frictionless Inclined Plane

Click For Summary
SUMMARY

The discussion focuses on calculating the tension in a cord connecting two boxes on frictionless inclined planes. The first box has a mass of 2.7 kg and is inclined at 28°, while the second box has a mass of 2.4 kg and is inclined at 64°. The correct approach involves applying Newton's second law, leading to the equations T - (2.7 kg)(9.8 m/s²)sin28 = -(2.7 kg)a and (2.4 kg)(9.8 m/s²)sin64 - T = -(2.4 kg)a. After correcting the angle from 68° to 64° and resolving the equations, the tension is calculated to be 17.4 N.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of trigonometric functions, specifically sine
  • Familiarity with inclined plane physics
  • Ability to solve linear equations
NEXT STEPS
  • Review the application of Newton's second law in multi-body systems
  • Study the effects of frictionless surfaces on motion
  • Learn about tension in cords and ropes in physics problems
  • Explore the use of free-body diagrams for analyzing forces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of tension calculations in multi-body systems.

ewiner
Messages
1
Reaction score
0

Homework Statement



Figure 5-60 shows a box of dirty money (mass m1 = 2.7 kg) on a frictionless plane inclined at angle θ1 = 28°. The box is connected via a cord of negligible mass to a box of laundered money (mass m2 = 2.4 kg) on a frictionless plane inclined at angle θ2 = 64°. The pulley is frictionless and has negligible mass. What is the tension in the cord?

Homework Equations



F = ma

T - mgsin\Theta = -ma

a = ( m / M + m )g

The Attempt at a Solution



Newton's 2nd law for each body:
m1: T - (2.7 kg)(9.8 m/s2)sin28 = -(2.7 kg)a
m2: (2.4 kg)(9.8 m/s2)sin68 - T = -(2.4 kg)a

Add m1 & m2: (2.4 kg)(9.8 m/s2)sin68 - (2.7 kg)(9.8 m/s2)sin28 = (2.4 + 2.7)a

Solve for a: a = 1.84

Insert a into m1: T-(2.7 kg)(9.8 m/s2)sin28 = -(2.7 kg)(1.84 m/s2)

T = 17.4 N

However, that is not the right answer... Where am I going wrong?
 

Attachments

  • fig05_68.gif
    fig05_68.gif
    2.2 KB · Views: 1,415
Physics news on Phys.org
Why do you have a negative sign in front of the masses? If the acceleration is negative it'll come out in your answer when you solve for a. You might be counting the negatives twice. Just my 2 cents.
 
Although you incorrectly threw in those negatives in front of the masses, you ignored that error and changed it correctly to a positive number when you did your equation additions and substitution to solve for T. Strange. But the other issue is that you copied the angle down incorrectly...the angle given is 64 degrees, but you used 68 degrees. Make that angle correction and solve for T using 2 significant figures in the result.
 
PhanthomJay said:
Although you incorrectly threw in those negatives in front of the masses, you ignored that error and changed it correctly to a positive number when you did your equation additions and substitution to solve for T. Strange.

Then he went back to the negative

Insert a into m1: T-(2.7 kg)(9.8 m/s2)sin28 = -(2.7 kg)(1.84 m/s2)
 

Similar threads

Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K