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http://www.xtremepapers.me/CIE/Internat ... _qp_43.pdf
Question 3 part (i)
Can some one please explain how to do it?
:Search:
Question 3 part (i)
Can some one please explain how to do it?
:Search:
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Maths Mechanics (M1) question!
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Method 1. Making the resultant the hypotenuse, we resolve it into horizontal and vertical components.
What's the vertical component equal to? Well there's T in the vertical part of the string, AND there's a component in the sloping part of the string.
So we end up with 3\sqrt{3}cos 30 = T + Tsin 30
So 9/2= 3T/2
And hence T = 3 N as required.
If we'd looked at the horizontal components we'd have.
3\sqrt{3}sin 30 = T cos 30
and again T=3 N.
The upshot of all this is it doesn't matter which way you do it. It depends which force(s) you want to resolve, the Tension (of which there are two parts) or the Resultant.
Hope that helps. I'll write method 2(resolving tension) if you don't understand it.
What's the vertical component equal to? Well there's T in the vertical part of the string, AND there's a component in the sloping part of the string.
So we end up with 3\sqrt{3}cos 30 = T + Tsin 30
So 9/2= 3T/2
And hence T = 3 N as required.
If we'd looked at the horizontal components we'd have.
3\sqrt{3}sin 30 = T cos 30
and again T=3 N.
The upshot of all this is it doesn't matter which way you do it. It depends which force(s) you want to resolve, the Tension (of which there are two parts) or the Resultant.
Hope that helps. I'll write method 2(resolving tension) if you don't understand it.
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Thanks for the detailed reply!, but I don't get how you get the angel 30? Is that angel PAN?
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The prism is in the form of a right angle triangle and as the angle of its slope with horizontal is 30 degree so the angle at which its perpendicular and hypotenuse meet is 60 degree. The force exerted on the pulley by string is halfway between them, so it makes an angle of 30 degree with both perpendicular and hypotenuse.
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ohh! ok Makes sense! but how are supposed to know that the resultant acts in the middle of the hypotenuse and perpendicular? Is it because this is a pulley?
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It's because of symmetry and fact.
You may want to know the other method of resolving tensions, that's easier than this one and you wouldn't need to bother about the 30 degree angle.
Method 2. Considering the tension. Lets call it T. We resolve the tension in the direction of where we know the resultant for will be (due to the symmetry) (and the resultant becomes the base in the triangle), and the vertical tension gives a component Tcos(30) in the direction of the resultant, and similarly the tension of string which supports particle on a slope.
Tcos(30) + Tcos(30) = 3\sqrt{3}
and again T = 3.
You may want to know the other method of resolving tensions, that's easier than this one and you wouldn't need to bother about the 30 degree angle.
Method 2. Considering the tension. Lets call it T. We resolve the tension in the direction of where we know the resultant for will be (due to the symmetry) (and the resultant becomes the base in the triangle), and the vertical tension gives a component Tcos(30) in the direction of the resultant, and similarly the tension of string which supports particle on a slope.
Tcos(30) + Tcos(30) = 3\sqrt{3}
and again T = 3.