# A man pulls his child in a sled at a constant velocity. The child and sled have a combined mass of 60 kg. The force applied on the handle is 200N at an angle of 15. Draw a free body diagram. What is the normal force acting on the sled?

## What is the force of friction acting on the sled?

Feb 28, 2017

With this kind of problem, it's very important that you keep track of your sign conventions. I set the y direction perpendicular (outwards from) to the ramp as positive and x direction parallel down the ramp as negative. Thus, $\vec{g} < 0$ in this convention.

Here's the free body diagram I came up with (treat the sled and child as one unified object): Remember that ${\vec{F}}_{g , | |} = m \vec{g} \sin \theta$ is the parallel force component due to gravity that opposes the pulling force and ${\vec{F}}_{g , \bot} = m \vec{g} \cos \theta$ is the perpendicular force component due to gravity that opposes the normal force.

Therefore, numerically, ${\vec{F}}_{p u l l}$ and ${\vec{F}}_{k}$ oppose each other, and numerically, ${\vec{F}}_{N}$ and $m \vec{g} \cos \theta$ oppose each other, and we can write a sum of the forces:

${\sum}_{i} {\vec{F}}_{x , i} = - {\vec{F}}_{k} + {\vec{F}}_{p u l l} + m g \sin \theta = 0$

${\sum}_{i} {\vec{F}}_{y , i} = {\vec{F}}_{N} + m g \cos \theta = 0$

where ${\vec{F}}_{k}$ is the reacting force of the kinetic friction, ${\vec{F}}_{p u l l} = \text{200 N}$ is the force used to pull the sled, and ${\vec{F}}_{g , | |} = m g \sin \theta$ is the force component of gravity parallel to the ramp. Note that ${\vec{a}}_{x} = 0$ due to the constant velocity.

The plus and minus signs in the equations are such that the forces numerically oppose each other when they add later on. I know that ${\vec{F}}_{N} > 0$ and ${\mu}_{k} > 0$, so ${\vec{F}}_{k} > 0$ (the minus sign accounts for left being negative). The reason why $m g \sin \theta$ in the same direction has a plus sign is because $\vec{g} < 0$ already.

You can see we can solve for the normal force right away.

$\textcolor{b l u e}{{\vec{F}}_{N}} = - m g \cos \theta$

$= - \left(\text{60 kg" xx -"9.81 m/s"^2 xx cos(15^@)) = color(blue)("568.54 N}\right)$

Once we have that, we should recall the definition of the kinetic friction force:

${\vec{F}}_{k} = {\mu}_{k} {\vec{F}}_{N}$,

where ${\mu}_{k}$ is the coefficient of kinetic friction.

Thus, the coefficient of kinetic friction is acquired like so:

$- {\vec{F}}_{k} + {\vec{F}}_{p u l l} + m g \sin \theta = 0$

$- {\vec{F}}_{k} = - {\vec{F}}_{p u l l} - m g \sin \theta = - {\mu}_{k} {\vec{F}}_{N}$

$\textcolor{b l u e}{{\mu}_{k}} = \frac{{\vec{F}}_{p u l l} + m g \sin \theta}{\vec{F}} _ N$

= (("200 N") + ("60 kg")(-"9.81 m/s"^2)sin(15^@))/("568.54 N")

$= \textcolor{b l u e}{0.0838}$

And the force of kinetic friction would be:

$\textcolor{b l u e}{{\vec{F}}_{k}} = {\vec{F}}_{p u l l} + m g \sin \theta$

= "200 N" + ("60 kg")(-"9.81 m/s"^2)(sin15^@)

$=$ $\textcolor{b l u e}{\text{47.66 N}}$