Question

A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and (b) and acceleration of 0.40 m/s2

Answer 1

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the  same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$