Newton's Third Law - Using pulleys and slopes

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SUMMARY

The discussion centers on a physics problem involving a 5.9 kg box on a frictionless 40-degree slope connected to a 2.1 kg hanging weight via a massless string over a massless pulley. The tension in the string when the box is held in place is calculated as 20.61 N. Upon release, the box will slide down the slope due to the greater mass of the box on the slope (m1). The tension in the string once the box begins to move requires solving two equations: -m1*g*sin(θ) + T = m1(-a) and m2*g - T = m2(a).

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of forces acting on inclined planes
  • Familiarity with tension in strings and pulleys
  • Basic trigonometry, specifically sine functions
NEXT STEPS
  • Learn how to derive equations of motion for objects on inclined planes
  • Study the concept of tension in systems with pulleys
  • Explore the use of free-body diagrams to analyze forces
  • Investigate the effects of friction on inclined planes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of tension and motion in pulley systems.

tastypotato
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Homework Statement


A 5.9 kg box is on a frictionless 40 degree slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.1 kg weight.

A.
What is the tension in the string if the 5.9 kg box is held in place, so that it cannot move?

B.
If the box is then released, which way will it move on the slope?

C.
What is the tension in the string once the box begins to move?

Homework Equations


Fnet=m*a


The Attempt at a Solution


A. The mass of the second box times the gravity will give me the solution to part A -> m2*(9.81)=20.61N
B. If the box is released then it will slide downwards on the slope because there is more mass in m1
C. This is where I'm stuck. My professor wrote down this equation for us to follow.

-m1*g*sin[tex]\theta[/tex]+T=m1(-a)
and
m2*g*T=m2(a)

but I don't know where to go from here.
 
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Hi tastypotato! :smile:

(have a theta: θ :wink:)
tastypotato said:
A 5.9 kg box is on a frictionless 40 degree slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.1 kg weight.

B.
If the box is then released, which way will it move on the slope?

C.
What is the tension in the string once the box begins to move?

B. If the box is released then it will slide downwards on the slope because there is more mass in m1

Sorry, but that's the wrong reason … "more mass" doesn't clinch it … what matter is the equation your professor wrote down for you …
C. …
-m1*g*sin[tex]\theta[/tex]+T=m1(-a)
and
m2*g*T=m2(a)

(and btw, this is https://www.physicsforums.com/library.php?do=view_item&itemid=26", not third)

These two equations are the Ftotal = ma equations for m1 and m2 separately

(you have to do them separately) …

the first one is of the components in the direction of the slope, and the second is for the vertical components …

a is the same in both equations because the string has fixed length, so if m1 goes distance x up the slope, m2 goes distance vertically down …

so i] can you prove those two equations are correct?

ii] eliminate T, and find a :smile:
 
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