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Is this a mistake on mechanics mark scheme ?
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I have calculated the forwards component parallel to the surface is :
z = 30 / cos(30) whereas they have done 30 * cos(30).
Am I right in saying that this is a mistake ?
Thanks.
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(Original post by mqb2766)
The component must be smaller than the original force
The component must be smaller than the original force
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#6
But how the 30/cos(30)? The 30 is the hypotenuse. The original force is always the hypotenuse, so the legs are always smaller, as you'd expect.
It would be better to draw the triangle with the forces/plane in the original orientation. Makes errors easier to find.
It would be better to draw the triangle with the forces/plane in the original orientation. Makes errors easier to find.
Last edited by mqb2766; 4 weeks ago
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#8
(Original post by seals2001)
How is it ? It acts horizontally , i.e parallel with the surface and you want the length of the force parallel with the slant ? Right ?
How is it ? It acts horizontally , i.e parallel with the surface and you want the length of the force parallel with the slant ? Right ?
See the previous post, which I was editing.
Last edited by mqb2766; 4 weeks ago
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(Original post by mqb2766)
No. The right angle is where the added leg hits the plane. The hypotenuse is the original horizontal force. The triangle isn't correct.
See the previous post, which I was editing.
No. The right angle is where the added leg hits the plane. The hypotenuse is the original horizontal force. The triangle isn't correct.
See the previous post, which I was editing.
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#10
(Original post by seals2001)
Ohhh, I get what you are saying now. So the components give an additive effect, you cant have negative components, so the force you are resolving into components has to be bigger than its individual components. Thanks
Ohhh, I get what you are saying now. So the components give an additive effect, you cant have negative components, so the force you are resolving into components has to be bigger than its individual components. Thanks
The tip about drawing the right angled triangle in the same orientation as the original parts is useful as it's easier to spot stuff like this. Your original triangle had the extra leg perpendicular to 30 so it's component would (in theory) be zero, but not according to your triangle. Also the extra leg was not perpendicular to the plane. It's harder to spot when you rotate the triangle.
Last edited by mqb2766; 4 weeks ago
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(Original post by mqb2766)
Basically yes (but independent - perpendicular components - pythagoras). You don't get extra force, compared to the original, by changing the orientation.
The tip about drawing the right angled triangle in the same orientation as the original parts is useful as it's easier to spot stuff like this. Your original triangle had the extra leg perpendicular to 30 so it's component would (in theory) be zero, but not according to your triangle. Also the extra leg was not perpendicular to the plane. It's harder to spot when you rotate the triangle.
Basically yes (but independent - perpendicular components - pythagoras). You don't get extra force, compared to the original, by changing the orientation.
The tip about drawing the right angled triangle in the same orientation as the original parts is useful as it's easier to spot stuff like this. Your original triangle had the extra leg perpendicular to 30 so it's component would (in theory) be zero, but not according to your triangle. Also the extra leg was not perpendicular to the plane. It's harder to spot when you rotate the triangle.
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