How Do You Calculate Tension and Work for a Sledge Pulled at an Angle?

  • Thread starter Thread starter cjknowland
  • Start date Start date
  • Tags Tags
    Force Friction
AI Thread Summary
To calculate the tension in the rope pulling a sledge at an angle, the normal force must be determined, which is affected by the angle of pull and the weight of the sledge. The key equation involves balancing forces, recognizing that the sledge moves at constant speed, indicating zero acceleration. The tension can be calculated using the formula T = mgμ / (cos(θ) + sin(θ)μ), where m is mass, g is gravitational acceleration, μ is the coefficient of friction, and θ is the angle of pull. Work done by the rope and mechanical energy lost due to friction can also be derived from these calculations. Understanding these relationships is crucial for solving the problem effectively.
cjknowland
Messages
6
Reaction score
0

Homework Statement




A sledge loaded with bricks has a total mass of 17.4 kg and is pulled at constant speed by a rope inclined at 19.2° above the horizontal. The sledge moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.500.

(a) What is the tension in the rope?

(b) How much work is done by the rope on the sledge?

(c) What is the mechanical energy lost due to friction?


Homework Equations





The Attempt at a Solution



I am having trouble finding the normal force with the pull at 19.2 degrees. If anyone could help in solving this problem it would be greatly appreciated.
 
Physics news on Phys.org
I believe the key is that the acceleration is zero but I do not know how to apply it...
 
F-fk=MA but in order to find fk, I need to use fk=(coefficient of friction)(normal force) but i do not know how to find the normal force with the information given.
 
okay I got an answer of 76.76... I don't know if its correct
 
Solving for T, I got 76.97 N

I'm willing to bet the discrepancy is rounding.
To be sure, I used
mg-Tsin(θ) = N
friction = (mg-Tsin(θ))μ
Tcos(θ)=friction = (mg-Tsin(θ))μ

T(cos(θ) + sin(θ)μ) = mgμ

so, T =mgμ/(cos(θ) + sin(θ)μ)
 
Last edited:
thank you so much =]
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top