Optimizing Ramp Length for Unicycle Climbing

[woaini]

Homework Statement

A person on a unicycle has a total weight of 900N. They need to climb to a height of one meter. If we build the ramp for a minimum pushing force of 50N, how long must the ramp be?

m=91.7kg
w=900N
F1=50N
Ry=1m
Rx=?
Rxy=?
θ=?

Homework Equations

.5mv²=mgh

F1=mgsinθ

The Attempt at a Solution

θ=arcsin(F1/mg)=3.18°

v=sqrt($\frac{mgh}{2m}$)=2.21m/s

I am not really sure at how to approach this problem. However, I have solved for the variables that I think are possible.

 

[NascentOxygen]

You found the angle of inclination correctly. All that is left is to find the dimensions of that triangle when its height is 1m.

energy ⇔ velocity does not come into this problem.

 

[Simon Bridge]

how much work to climb 1m?
how much force is available?
What is the relationship between work, force, and distance?

 

[woaini]

You found the angle of inclination correctly. All that is left is to find the dimensions of that triangle when its height is 1m.

energy ⇔ velocity does not come into this problem.

The answer that I get is 18m.

1/sin(3.18)=18m

Do I have to account for circular acceleration of the unicycle?

 

[woaini]

how much work to climb 1m?
how much force is available?
What is the relationship between work, force, and distance?

The work to climb 1m is W=Fd .

The amount of available force to push is a minimum of 50N.

Therefore the slanted distance of the ramp is unknown and the work needed to be done is unknown. What is known is the minimum force required which is 50N. I do not see how this information can be relevant to the question. Can you please explain further?

 

[nasu]

You know the work needed to be done. You need to raise a 900N weight by 1m.
The same work can be done either by applying a 900N force over a distance of 1m or a 50N force over ...

 

[woaini]

You know the work needed to be done. You need to raise a 900N weight by 1m.
The same work can be done either by applying a 900N force over a distance of 1m or a 50N force over ...

So essentially you can set W=W.

Which would therefore be Fd=Fd and 900(1)=50d and d=18m.

I also solved this by finding the angle and doing simple trigonometry to find 18m.

So is this answer right, or am I doing something wrong?

 

[Simon Bridge]

You verified your previous answer by another method - which means you don;t need us to tell you if you were right of not :) Most people find the work-energy theorem to be simpler than trig.

Strictly speaking - some energy goes to making the unicycle wheel move. But you have not been given information to do that bit so it won't be expected.

 

[woaini]

Just making sure. :)

And it does make sense that I cannot calculate how the unicycle moves since I don't have radius or time.