Magnitude of the Force of Friction

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Homework Help Overview

The problem involves a sled being pulled at a constant velocity across a horizontal snow surface, with a specified force applied at an angle. The objective is to determine the magnitude of the force of friction acting on the sled.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force and the frictional force, noting that the net horizontal force must be zero due to the constant velocity. There are attempts to clarify the components of the applied force using trigonometric functions.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying the relationships between the forces involved. Some guidance has been offered regarding the use of trigonometric relationships to find the normal force and frictional force.

Contextual Notes

Participants are navigating the implications of the sled moving at constant velocity, which suggests that the forces are balanced. There is also a mention of using the coefficient of friction in relation to the normal force, indicating a potential area of exploration.

becsantos
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Homework Statement


A sled is pulled at a constant velocity across a hrizontal snow surface. If a force of 8.0 x 10^1 N is being applied to the sled rope at an angle of 53 degrees to the ground, what is the magnitude of the force of friction of the snow acting on the sled?


Homework Equations





The Attempt at a Solution


I made a right triangle with one angle being 53 degrees and the hypotenus as 80 N.
80sin(53)= 63.9 ; horizontal axis
80cos(53)= 48.14 ; vertical axis
 
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hi,
since there is no acceleration, thus the net horizontal force must be zero.
then you will get your frictional force.
 
welcome to pf!

hi becsantos! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
becsantos said:
I made a right triangle with one angle being 53 degrees and the hypotenus as 80 N.
80sin(53)= 63.9 ; horizontal axis
80cos(53)= 48.14 ; vertical axis

53° is the angle to the ground, so the horizontal side of the triangle is 80cos53° :smile:

(cos = adj/hyp)
 
What equation am I using to find frictional force?
Ff=UkxN ? ow do I solve?
 
hi becsantos! :smile:

(btw, no need to send a pm … anyone who posts in a thread gets automatic email notification of all new posts :wink:)

no equation needed, just use the triangle …

that will give you both the normal force and the friction force :smile: (and if you do want to find µk, then you can use Ff = µkN)
 
which is the frictional force?
 
oh come on :rolleyes:

what do you think?​
 
horizontal, right?
 
yes of course …

the surface is horizontal, so the friction must be horizontal

ok, so what is the magnitude of the force of friction of the snow acting on the sled?
 

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