M1 Moments Watch

StylaBoy
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#1
Report Thread starter 7 years ago
#1
Ok, I'm stuck on Moments:rolleyes:


First of all, I thought you can only take moments when the distance is perpendicular to the force. I've highlighted it in the picture and that does not seem to be perp. ?

And 2ndly, can you tell me when we take sin and when cosine when measuring angles? Because in this example, I don't know why they do sin and not cosine for instance:confused:

many thx in advance
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ttoby
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Report 7 years ago
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You're right that the distance isn't perpendicular to the force, but there's a way of getting round this.

If you look at the 8N force, you'll see that they've drawn a long, straight arrow in the direction where it acts. When dealing with moments, you're allowed to consider the 8N force as acting anywhere along that arrow. So we could, for instance, 'move' the force along the arrow so that it's in a position where the distance from P to the force is perpendicular to the arrow.

This is what I've done in the picture below. I've 'moved' the force along to where the end of the grey line is then the grey line represents the perpendicular distance.



Now as for sines and cosines, this diagram might help:



You can check it's valid by using the standard definitions of sin and cos, but when you're doing M1 it's helpful to think of the components of a force etc like this.
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ilyking
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Report 7 years ago
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you have to realize that the 8N force has some vertical (and horizotal) affect, it's logical when you think about it, try not to think like a robot if you know what I mean, especially with mechanics
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StylaBoy
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#4
Report Thread starter 7 years ago
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Okay, thanks for that, but I simply don't get the 2nd answer..

With all respect, I know you are trying to help me, but I don't know what to do with the 2nd picture. It doesn't say anything to me.

Cheers
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ttoby
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Report 7 years ago
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(Original post by StylaBoy)
Okay, thanks for that, but I simply don't get the 2nd answer..

With all respect, I know you are trying to help me, but I don't know what to do with the 2nd picture. It doesn't say anything to me.

Cheers
First of all, we can check it's valid mathematically. Opposite/hypoteneuse = (x sin t)/x=sin t and adjacent/hypoteneuse = (x cos t)/x = cos t

The way to use it is take some quantity with magnitude and direction, such as a force or the length and direction of a piece of string.

We want to find out its horizontal and vertical components. If the magnitude is x and the force is directed t degrees above the horizontal then the horizontal component will be x cos t as shown in the diagram, and the vertical component will be x sin t.

This is handy for resolving forces for example. Another way of using it would be to find the component of a force parallel and perpendicular to the plane.

Here are some examples:

The force F has horizontal component F cos t and vertical component F sin t

The force has parallel component 6 cos t and vertical component 6 sin t

And in the question you asked about, the distance from P to the force is 2 sin 50.
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