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So there was a this question I have been wondering about mechanics. Consider an inclined plane of angle thetha with mass $M$ and an object sliding on it of mass $m$.The inclined plane is in a weighing machine.What weight will the machine give while the mass $m$ is sliding how it compares to when it isn't sliding.

I was thinking about the acceleration of the center of mass and the external forces,then clear for the Normal force of the mass $M$ but I have trouble with the vector form of acceleration cause it isn´t only in the y direction so how can it equal the external forces that are only in $y$ ($F=ma$)? Or is there another way to get the Normal Force? First I thought is equal to the weight as normal but I don't think it's that easy. So thanks for the help and ideas.

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Consider a "weighing machine" as simply a force gauge that measures force perpendicular to a surface that is applied by the surface, or the normal force. In this case, it is measuring the force applied by the ground to the inclined plane. The two key principles we need are Newton's 3rd and Newton's 2nd laws. Let's examine the first scenario, all objects are stationary. What are the forces acting on each element of the system? Well for the block there is gravity pointing downward, which is opposed perpendicularly to the surface of the plane by the normal force of the plane on the block, and opposed parallel to the incline by (let's say) friction. Now what are the forces acting on the incline: naturally the equal and opposites to the normal and friction forces applied by the plane on the block. Additionally, there is gravity acting (downwards) and the normal force of he ground on the incline (this is the particular force we are measuring). (Note If we constrain the incline to not move horizontally, there must also be some "friction" force applied to the incline, but as it is perpendicular to the force we are measuring, it's doesn't really need to be considered). When everything is at rest these forces sum to 0 for each object.

Now let's examine the case where the block is sliding (let's say without friction). Thus the opposite friction force acting on the incline is absent. As we still assume the incline is stationary (or most importantly, stationary vertically) our normal force applied by the ground on the incline must change to meet the requirement that the net force on the incline is 0.

To explicitly carry our the calculation I would recommend drawing out a free body diagram and summing the forces accordingly. Hope this helps.

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  • $\begingroup$ Yeah the net normal acceleration would be g times the cosine of thetha but then the weighing machine would measure this force and the weight of the inclined plane? Is that this force is perpendicular to the plane, but not to the weighing machine so it contributes with only one part? $\endgroup$ – Rafa Flores May 19 '17 at 1:36
  • $\begingroup$ @RafaFlores My mistake, I misread your question, I'll edit my answer accordingly. $\endgroup$ – Andrew Murphy May 19 '17 at 1:49
  • $\begingroup$ Thanks I think I solve it my answer is that the Normal force is mass M times g plus mass m times g times the cosine square of thetha $\endgroup$ – Rafa Flores May 19 '17 at 3:41
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The wedge has no component of acceleration in the vertical direction, so this weighs $Mg$. The object sliding down the wedge has a component of acceleration $a$ in the vertical downward direction; its weight is $m(g-a)$. Note that if $a=g$ then this object is in free fall so it weighs nothing.

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