You are trying to raise a bicycle wheel of mass m and radius R up over a curb of

Question

You are trying to raise a bicycle wheel of mass m and radiusR up over a curb of height h. To do this, you apply a horizontal force F? . (Figure 1)

Part A

What is the least magnitude of the force F?  that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel?

Part B

What is the least magnitude of the force F?  that will succeed in raising the wheel onto the curb when the force is applied at the top of the wheel?

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Part C

In which case is less force required?

In which case is less force required?

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Explanation / Answer

This is just pure TRIGONOMETRY.. .but it's hard to solve in here.. coz I can't draw diagrams...

I'll show you the maths...

Let's draw your diagram....
Draw your diagram... Circle, with a curb of height h...
a) From the centre of the wheel, draw a radius straight down... representing r, and mg as a force. Call it CB
b) From the centre draw a radius to the curb height. Call it CH
c) From the Centre, draw a horizontal line to represent your Force applied... call it CF

Label the angle at the centre, from the bottom as alpha (CB-CH), and then theta next to it (CH-CF)

Now the maths.... Moments of Inerta... with the curb being the pivot, or fulcrum
1. the moment weight of the wheel...
mgsin(alpha) * r
2. the moment of the force
Fsin(theta) * r

For the minimum force.. these 2 must equal.
mgsin(alpha) * r = Fsin(theta) * r ---> (r's cancel)
..mgsin(alpha) = Fsin(theta)

But... theta = 90 - alpha
..mgsin(alpha) = Fsin(90-alpha)
..mgsin(alpha) = Fcos(alpha)
..mgtan(alpha) = F

Now the angle alpha can be easily calculated.
..cos(alpha) = (r-h)/r
therefore: alpha = cos^(-1)[(r-h)/r]
hence
F = mg tan{cos^(-1)[(r-h)/r]}



The second part.. the trig is a bit harder...
Draw your diagram... Circle, with a curb of height h...

a) This time... draw a vertical diameter from top to bottom of your wheel. Call this TCB where C is the centre.
b) From C draw a radius to the top of the curb, call it CH
c) At point T (top of wheel) draw a horizontal line to represent F applied. Call it TF
d) Draw a chord (TH) from point T to point H. (this is the lever, and point H is the pivot, or fulcrum)

Label you angles... at T, top of wheel...again with the bottom angle as (alpha) TC-TH, and (thetha) next to it TH-TF

Now the Maths for the MOMENTS should be the same as the first part... except you don't multiply the forces by r, but rather by the chord TH. The lever chord cancels out anyhow.. leaving:
..mgtan(alpha) = F

But this time, alpha is different to before...
Label angle at the centre of the wheel... CB-CH as (beta)
Now from part 1. it follows that...
cos(beta) = (r-h)/r
and... beta = cos^(-1)[(r-h)/r]

But as you can see.. the triangle CTH is an isoceles triangle, with beta being the exterior angle of this triangle... therefore:
beta = 2 alpha
alpha = (1/2) beta
..alpha = (1/2) cos^(-1)[(r-h)/r]

So
F = mg tan{(1/2)cos^(-1)[(r-h)/r]

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