Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to the horizontal ground, and the sled is being pulled at a constant speed. If the sled and snow have a total mass of m, the acceleration due to gravity is g, the magnitude of the normal force is FN, and the coefficient of kinetic friction between the sled and the ground is μk, what is the angle θ that the rope makes with the ground in terms of these five quantities? (Use the following as necessary: m, g, FN, and μk.)

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nuuk nuuk · Answer 1 · 2019-10-16T06:46:16+02:00

Answer:

Explanation:

Given

rope makes an angle of [tex]\theta [/tex]

Mass of sled and snow is m

Normal Force [tex]=F_N[/tex]

applied Force is F

as Force is pulling in nature therefore normal reaction is given by

[tex]F_N=mg-F\sin \theta [/tex]

Also [tex]F\cos \theta =f_r[/tex]

[tex]f_r=\mu _k\cdot F_N[/tex]

[tex]f_r=\mu _k\cdot (mg-F\sin \theta )[/tex]

[tex]F\cos \theta =\mu _kF_N[/tex]-------1

[tex]F\sin \theta =mg-F_N[/tex]---------2

Squaring 1 & 2 and then adding

[tex]F^2=(\mu _kF_N)^2+(mg-F_N)^2[/tex]

[tex]F=\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}[/tex]

Substitute value of F in 1

[tex]cos\theta =\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}}[/tex]

[tex]\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})[/tex]