Question

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a man drags a 10.0 kg bag of mulch at a constant speed, applying a 22.5 n force at 32.0 degrees. what is the friction force acting on the bag?

Answer 1

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$, we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

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