Calculate the moment when the pedal

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The discussion revolves around calculating the moment generated by a bicycle pedal under a force of 350 N at two angles: horizontal and 65º below horizontal. The moment is defined as the product of force and the perpendicular distance from the pivot, represented by the equation M=Fxd. When the pedal is horizontal, the moment is straightforward, but at an angle, the perpendicular distance must be calculated using the sine of the angle. Confusion arises regarding the angles and the correct application of the moment formula, particularly distinguishing between the angle of rotation and the angle used in the sine function. Ultimately, the conversation emphasizes the importance of understanding the relationship between force, distance, and angle in calculating moments accurately.
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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.
 
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wally25feb said:
Could someone solve this question please?
That would be against the rule - but we can help you to solve it.
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.
So what is the definition of a "moment"?
 
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
 
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.
 
Sorry, it's 65 deg. My bad. Then 25 degree. Does it make sense now?
 
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?
 
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.
 
wally25feb said:
The 350N force is tangent to each point of the circle
Not so. The 350N force is always straight down (no cleats, I guess).
 
M=Fxdxsinθ
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].
 
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