Question

A boy drags a 100 N sled up a 20 slope at constant velocity. If the coefficient of friction between sled and hill is 0.20, what force must he exert at an angle of 35 with respect to the hill?

Answer 1

Answer:

$F = 56.75 N$

Explanation:

As per the free body diagram of the box we can say that the applied force must be greater than or equal to the sum of friction force and component of the weight of the box along the hill

So we can say that

$F_{net} = F cos35 - (mgsin20 + F_f)$

on the other side the force perpendicular to the plane must be balanced so that it remains in equilibrium in that direction

so we can say that

$F_n + Fsin35 = mgcos20$

now we will have

$F_f = \mu F_n$

$F_f = (0.20)(mg cos20 - Fsin35)$

now we have

$0 = Fcos35 - mg sin20 - (0.20)(mg cos20 - Fsin35)$

$F(cos35 + 0.20 sin35) = mg sin20 + 0.20 mgcos20$

$F = \frac{mg sin20 + 0.20 mgcos20}{(cos35 + 0.20 sin35)}$

$F = \frac{100(sin20 + 0.20 cos20)}{(cos35 + 0.20 sin35)}$

$F = 56.75 N$