Question

a 50 kg sled is pulled by a rope at an angle of 30 degrees above the horizontal with a force of 35 N . this is sufficient to keep the sled moving at a constant speed. What is the coefficient of sliding friction for the sled-snow surface?

Answer 1

A $50\mathrm{kg}$ sled is pulled by a rope at an angle of 30 degrees above the horizontal with a force of 35 N. this is sufficient to keep the sled moving at a constant speed. What is the coefficient of sliding friction for the sled-snow surface?

Solution:

Such as the sled moving at a constant speed the force $F$ is equal to sum of forces $F_s$ and $F_f$ , were $F_f - f$ frictional force

$$F = F _ {s} + F _ {f}$$ $$F _ {f} = k N$$ $$k N = F - F _ {s}$$ $$\boldsymbol {k} = \frac {F - F _ {s}}{N} = \frac {F - m g s i n 3 0 {}^ {\circ}}{m g c o s 3 0 {}^ {\circ}}$$ $$\boldsymbol {k} = \frac {3 5 - 5 0 * 9 . 8 * \sin 3 0 {}^ {\circ}}{5 0 * 9 . 8 * \cos 3 0 {}^ {\circ}} = - 0. 4 9$$

The sign “-” means that friction force directed to an opposite side to F

Answer: 0.49