Amount of force to pull a sled up an icy slope

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Homework Help Overview

The problem involves calculating the force required to pull a sled up an icy slope at a constant velocity, considering the effects of friction and gravity. The sled has a mass of 19 kg and a coefficient of kinetic friction of 0.050. The discussion has shifted to analyzing the scenario when the sled is on a slope of 30 degrees above the horizontal.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for a free body diagram (FBD) to visualize forces acting on the sled. There is an exploration of how to calculate the normal force on a slope and the relationship between the normal force and friction. Questions arise about the forces that need to be overcome when pulling the sled, including the tension in the rope and gravitational force.

Discussion Status

The discussion is active, with participants providing guidance on the importance of the normal force and its calculation. There is a focus on understanding the components of forces acting on the sled, particularly in the context of a slope. Multiple interpretations of the forces involved are being explored, and participants are questioning assumptions about the relationship between weight and friction.

Contextual Notes

Participants note the need to consider the angle of the slope and its effect on the normal force and friction. There is an emphasis on checking units and ensuring proper application of physics principles in the context of the problem.

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Homework Statement



You're using a rope to pull a sled at constant velocity across level snow,with coefficient of kinetic friction 0.050 between sled and the snow. The sled is loaded with physics books, giving a total mass of 19 {\rm kg}.

Homework Equations



F=\mu*n

The Attempt at a Solution



I tried solving for the force and the force of friction separately then subtracting between them... Not sure if that's correct.
EDIT: I'm sorry I'm on the second part of this question which asks for the same force but at a slope of 30 degrees above horizontal.
 
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Okay, a free body diagram (FBD) would be useful here, if you haven't yet drawn it.

Remember the equation for friction involves the normal force, not the weight. What would be the normal force when the sled is on a slope?

When you have the friction force, what other force is there that you need to overcome if you are pulling the sled up a hill?
 
The normal force would just be cos30*g right? Since weights not a factor in the friction...?

The other force beside the frictional is the tension of the rope being pulled which would factor in m*g and cos30?
 
Capncanada said:
The normal force would just be cos30*g right? Since weights not a factor in the friction...?

The other force beside the frictional is the tension of the rope being pulled which would factor in m*g and cos30?

I guess my statement came out a little confusing. Weight is a factor in friction only to the point that it contributes to the normal force. So the normal force would be

F_N = m g\cos{(30)}

Remember to check your units - g is an acceleration. It needs a mass to become a force.

And you are applying the tension, but what force are you overcoming? The tension you provided would overcome this force.

Take a look at your FBD. You should have the friction, the normal force, the tension on the rope and one other force.
 
The final force being gravity
 
hi

if the applied force is 30 degrees above the horizontal then the upward component would be
F_{app}\sin(30). now in FBD , in the vertical direction there are 3 forces, normal force N is acting on the sled upward , F_{app}\sin(30) is acting on the sled upward and force due to gravity , mg , is acting on the sled downwards. since there is no net force in vertical direction,

F_{y-net}=0

use this to solve for the normal force N.
 

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