A boy pulls a sled along a horizontal road, applying a force of 50N to them at an angle of 30 degrees to the horizon.

A boy pulls a sled along a horizontal road, applying a force of 50N to them at an angle of 30 degrees to the horizon. What is the frictional force between the sled and the road if the sled moves evenly?

F = 50 N.

∠α = 30 °.

a = 0 m / s2.

Ftr -?

Let us write Newton’s 2 law for sledges in vector form: m * a = F + m * g + N + Ftr, where F is the force with which the sled is pulled, m * g is gravity, N is the surface reaction force, Ftr is the force friction.

Let us find expression 2 of Newton’s law for projections onto the coordinate axes.

ОХ: 0 = F * cosα – Ftr.

OU: 0 = F * sinα – m * g + N.

Ftr = F * cosα.

The friction force Ffr is balanced by the projection of the force F with which the sled is pulled.

Ftr = 50N * cos30 ° = 43.3 N.

Answer: when moving on the sled, the friction force Ftr = 43.3 N.