# Homework Help: Calculate the moment when the pedal

1. Feb 22, 2013

### wally25feb

Could someone solve this question please?

The pedal of a bicycle is 18 cm long. When it is pushed downwards by a cyclist it experiences a force of 350 N. Calculate the moment when the pedal is horizontal and then when it is 65º below the horizontal.

Thanks.

2. Feb 22, 2013

### Simon Bridge

That would be against the rule - but we can help you to solve it.
So what is the definition of a "moment"?

3. Feb 22, 2013

### wally25feb

Hi Simon. Thanks for replying. To cut to the chase,

For horizontal positions:

M=FXd=-350x0.18=-63

For 60 degree position:

M=FXdxsin60=-63xsin60

Am I right?

Thanks

4. Feb 22, 2013

### Simon Bridge

Please show your reasoning... why no angle in the first one and a sin60 on the second one?

Isn't the angle on the second one 65degress?
Isn't this angle below the horizontal?
It helps to sketch the situation.

5. Feb 22, 2013

### wally25feb

Sorry, it's 65 deg. My bad. Then 25 degree. Does it make sense now?

6. Feb 22, 2013

### Simon Bridge

It's 65deg the 25deg??? No - doesn't make sense.
What are you doing with these numbers?
Did you sketch the situation and resolve the force into radial and tangential components? Or did you work the problem by the perpendicular distance from the axis to the line of action?

7. Feb 22, 2013

### wally25feb

I think I have found the solution. Have a look Simon.

Alright, let me put everything in the right order. Moment is the turning effect of a force around a pivot when the applied force is at right angle with the distance from the pivot, it can found by using the following equation:

M=Fxd

However, when the distance is not at right angle with the exerted force, then the perpendicular distance is equal to dxsinθ. Hence, the moment in this case can be calculated as:

M=Fxdxsinθ

I think this is my source of confusion but reading your comment made me reach to another conclusion. The pedal is making a circle motion about the pivot and its distance (18cm) is the radius to the circle. The 350N force is tangent to each point of the circle, resulting the same moment in each point as long as there is no change in 350N force because the radius of a circle is always orthogonal to the tangent at each touching point.

8. Feb 22, 2013

### haruspex

Not so. The 350N force is always straight down (no cleats, I guess).

9. Feb 22, 2013

### Simon Bridge

Which angle in this theta supposed to be?
Hint: it isn't always the rotation angle of the system.

BTW: you can leave off the multiplication symbol ... it's too confusing.
i.e. the moment is $\vec{M}=\vec{F}\times\vec{d}$ and it has magnitude: $M=Fd\sin\theta$ where $\theta$ is the angle between... [complete the sentence].