Magnitude of the Force of Friction

[becsantos]

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

 

[weesiang_loke]

hi,
since there is no acceleration, thus the net horizontal force must be zero.
then you will get your frictional force.

 

[tiny-tim]

welcome to pf!

hi becsantos! welcome to pf!

(try using the X² icon just above the Reply box )

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°

(cos = adj/hyp)

 

[becsantos]

What equation am I using to find frictional force?
Ff=UkxN ? ow do I solve?

 

[tiny-tim]

hi becsantos!

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

no equation needed, just use the triangle …

that will give you both the normal force and the friction force (and if you do want to find µ_k, then you can use F_f = µ_kN)

 

[becsantos]

which is the frictional force?

 

[tiny-tim]

oh come on …

what do you think?​

 

[becsantos]

horizontal, right?

 

[tiny-tim]

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?