# A skier goes down a slope -- Ignoring friction, calculate the acceleration

## Homework Statement:

A skier goes down a slope of 58% (30º). Ignoring friction, calculate:
a) The ##x## and ##y## components of the acceleration (parallel and perpendicular to the slope)
b) If it starts from repose, calculate the velocity it will have within one minute
c) Draw a diagram with all the vectors that play part in the exercise, and name them properly

## Relevant Equations:

Newton's second law
Basic kinematics
a) ##a_y=\dfrac{\sum{F_y}}{m}=\dfrac{N-mg\cos{\alpha}}{m}=(1-\cos{\alpha})g##
##a_x=\dfrac{\sum{F_x}}{m}=\dfrac{mg\sin{\alpha}}{m}=g\sin{\alpha}##
##a_y=(1-0,866)9,81\;m/s^2=1,31\;m/s^2##
##a_x=(0,5)9,81\;m/s^2=4,91\;m/s^2##
How can it be a perpendicular acceleration?; which coordinate system am I using?. I've solved it without no clue, just Newton's 2nd law.
b)##v_x=a_x\cdot{t}=4,91\;m/s^2\cdot{60\;s}=295\;m/s##
##v_y=a_y\cdot{t}=1,31\;m/s^2\cdot{60\;s}=78,6\;m/s##
##v=\sqrt{8,70\cdot{10^4\;m^2/s^2}+0,618\cdot{10^4\;m^2/s^2}}=305\;m/s##
Does it make sense?
c) I've found a picture (the first shown in the link's article), but I think there are too many vectors. The only vectors that play part, from my point of view, are ##\vec{N}## and weight

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kuruman
Homework Helper
Gold Member
How can it be a perpendicular acceleration?; which coordinate system am I using?. I've solved it without no clue, just Newton's 2nd law.
It looks like you are using a coordinate system where the x-axis is along the incline and the y-axis perpendicular to the incline. As for why there is a perpendicular acceleration, ask yourself the question "does the skier fly off or sink into the slope?" If the answer is no, then the skier's velocity is not changing in the perpendicular direction, is it?

The diagram that you referenced does not really show too many vectors. The arrows shown in red are the actual forces acting on the block. This agrees with the diagram that you are considering. The blue arrows are the two components of the weight ##\vec G##. Strictly speaking, the diagram should show either ##\vec G## or the pair ##\vec F_1## and ##\vec F_2##.

Last edited:
• mcastillo356
Thanks. But, which is the perpendicular component of the acceleration?##\vec{0}##? The exercise is part of an exam I didn't pass.

kuruman
Homework Helper
Gold Member
Yes, the perpendicular component of the acceleration is zero. In the perpendicular direction the normal force away from the incline and the component of the weight into the incline are equal in magnitude. If they were not, the skier's velocity would change in the perpendicular direction. That means the skier would acquire a velocity either into the slope and sink in the snow or away from the slope and lose contact with the slope. This does not happen, therefore we must conclude that the acceleration component perpendicular to the slope is zero.

When you apply Newton's second law to force diagrams, it pays to be systematic as follows.
1. Add all the forces acting on the system as vectors.
Here in the perpendicular y-direction we have, from the diagram ##F_{net,y}=N-mg\cos\alpha##. In the parallel x-direction ##F_{net,x}=mg\sin\alpha##.

2. Declare the acceleration vector on the basis of additional information about the motion of the system.
Here the acceleration vector is entirely along the incline (x-direction) because of the additional information that the skier does not sink into or fly off the incline, i.e ##a_y=0## and ##a_x \neq 0##.

3. Assemble Newton's law equation in the two directions
##F_{net,x}=ma_x~\rightarrow~mg\sin\alpha=ma_x.##
##F_{net,y}=ma_y~\rightarrow~N-mg\cos\alpha=0.##

See how it works?

• mcastillo356
Fantastic!. Now everything is logic. Salutes, and thanks again!

• kuruman
kuruman
•
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