Question

Newton's second law: Rudolph the red nosed reindeer is pulling a 25 kg sled across the snow in a field. The coefficient of kinetic friction is .12 The rope that is pulling the sled is coming off at a 29 degree angle above the horizontal. Find the force in the rope when the acceleration is .12 m/s^2.

Answer 1

Answer:

$F=39,68N$

Explanation:

Data:

Mass $m=25 Kg$

Coefficient of kinetic friction $\mu=0.12$

Angle = $29^{0}$

Acceleration = $0.12 \frac{m}{s^{2} }$

Solution:

By Newton's first law we know that for the x-axis:

$F_{rope_x}-F_f=F_R$ Where $F_R$ is the resulting force, and $F_f$ is the friction force.

And for the y-axis:

$F_{rope_y}+N=W$, where N is the normal force, and W is the weight of the sled.  

We know that the resulting force's acceleration is $0.12 \frac{m}{s^{2} }$, and by using Newton's second law, we obtain:

$F=m.a$

$F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N$ .

Now, the horizontal component of the force in the rope will be given by

$F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f$, since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.

To obtain the friction force, we must know the normal force:

$F_f=\mu. N$

Clearing N in the y-axis equation:

$N=W-F_{rope_y}=W-F_{rope}.sin(29^0)$

So we can express the x-axis equation as follows:

$F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))$

Finally, solving for F we get

$F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))$

$F=39,68N$