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M1/M2 question

When figuring out a question about a particle sliding down an inclined plane, is the downwards component of the weight always mgcosx? Or does it change?
downwoards as in the slope ? and what angle is X
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Original post by CookieGhoul
downwoards as in the slope ? and what angle is X


Well the reason I ask is because of this question:

A box of mass 2kg is projected with speed 6 m/s up a line of greatest slope of a rough place inclined to the horizontal. The coefficient of friction between the box and the plane is 1/3. Use the work -energy principle to calculate the distance the box travels up the plane before first coming to rest.

I know how to do it, the only problem im encountering is to find what the reaction force is. :frown:
Original post by T13
Well the reason I ask is because of this question:

A box of mass 2kg is projected with speed 6 m/s up a line of greatest slope of a rough place inclined to the horizontal. The coefficient of friction between the box and the plane is 1/3. Use the work -energy principle to calculate the distance the box travels up the plane before first coming to rest.

I know how to do it, the only problem im encountering is to find what the reaction force is. :frown:


Right. If you draw a free-body force diagram, things will appear more easier. The most important thing to do in a question like this, or in mainly all questions in mechanics, is to draw a diagram, and label the forces.

Try to resolve parralel to the plane and perpendicular to the plane
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Original post by CookieGhoul
Right. If you draw a free-body force diagram, things will appear more easier. The most important thing to do in a question like this, or in mainly all questions in mechanics, is to draw a diagram, and label the forces.

Try to resolve parralel to the plane and perpendicular to the plane


I did, and I seem to get that the component that's perpendicular to the plane is 2gcos30, but thats because our teacher told us that the component thats perpendicular is always cosx
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I did, and I seem to get that the component that's perpendicular to the plane is 2gcos30, but thats because our teacher told us that the component thats perpendicular is always cosx


Right. Thats correct.So whats the issue now ?
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Original post by CookieGhoul
Right. Thats correct.So whats the issue now ?


oh, because i kept getting the answer wrong and I checked the mark scheme and it said that the reaction force was 2gsin30, thats what confused me :redface:
Which angle is 30 ? Is it the angle of the inclined plane to the horizontal ?
Like this ?


Is it A, B, C ?
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Original post by CookieGhoul
Which angle is 30 ? Is it the angle of the inclined plane to the horizontal ?
Like this ?


Is it A, B, C ?


B? I really am not sure :s-smilie:
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Original post by CookieGhoul
Which angle is 30 ? Is it the angle of the inclined plane to the horizontal ?
Like this ?


Is it A, B, C ?


B? Im really not sure :s-smilie:
If its B, then obviously it will be the other way around. Cos when they say that the normal reaction is mg cos 30, the 30 degrees is the angle from the horizontal to the inclined plane, that means angle A

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