Tension of 2 Bodies on a Frictionless Inclined Plane

AI Thread Summary
The problem involves two boxes on frictionless inclined planes connected by a cord, requiring the calculation of tension in the cord. The equations derived from Newton's second law for each mass lead to a system that can be solved for acceleration and tension. An error was identified in the angle used for one of the masses, which should be 64 degrees instead of 68 degrees. Additionally, there was confusion regarding the use of negative signs in the equations, which could lead to incorrect results. The correct tension, after addressing these issues, should be calculated with attention to significant figures.
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: 1,407
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)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top