# Normal component of force exerted

Suppose a simple setup where a free load (red) is supported by an inclined plane (pink) and a block at the end of the inclined plane (blue) touching the load as shown:

Is the force normal to the pink plane exerted by the red load smaller than that of the case where the load doesn't make contact with the blue block (assuming the load can somehow stick to the plane)? I just wonder how it'd affect the reading on a scale if it was positioned on the pink plane. My take is that the normal force would be the same (thus, the scale readings as well) regardless of the blue block because, since the force exerted normal to the plane is $$m \times g \times cos(\theta)$$, $$m$$, $$g$$ and $$\theta$$ are variables independent of the blue block. Is that correct?

• Yes, so let me check if I made a mistake
– user65081
Jul 12 '20 at 19:17
• @Wolphramjonny If so, then the resisting force would just be the reverse of the force exerted due to gravity? Jul 12 '20 at 19:19
• No, the sum of the three forces, gravity, N and F (the new horizontal force), must be zero. So N does change
– user65081
Jul 12 '20 at 19:21
• @Wolphramjonny Will there be a new vertical force though? Jul 12 '20 at 19:23
• Added to the answer
– user65081
Jul 12 '20 at 19:27

## 2 Answers

On the axis perpendicular to the plane you now have $$N-mg \cos \theta-F\sin \theta=0$$, which changes $$N$$.

I assumed F is horizontal, but it might not be depending on the configuration. But as far as there is a horizontal component then N will increase

• I see. So under what configuration will there be only a horizontal component? If the inclination was >0 and <90 degrees, then there should be a vertical component, right? Jul 12 '20 at 19:31
• if the load is spherical and touches the block with a normal angle that is horizontal then F is horizontal, any other angle will have a vertical component
– user65081
Jul 12 '20 at 19:33
• Right. In the case where $F$ is purely horizontal, is it the negative of the horizontal component of $mg$, i.e. $F=\frac{mg}{tan(\theta)}$? If so, $N=2mg \cos(\theta)$. That doesn't look quite right. Jul 12 '20 at 19:48
• $F=mg\tan\theta$
– user65081
Jul 12 '20 at 19:57

Yes, that's correct. But if blue block and the red mass are in contact and now the red mass is stationary there is a force supplied by the blue block on the red mass. That must be considered as well that force is probably horizontal so it has a component that is perpendicular to the plane. This force will add to the normal force on the plane on the red block. This will, as @Wolphramjonny stated, in his comment, increase the weight shown on the scale.

• Is there just a new horizontal force? Will there be a new vertical force? Jul 12 '20 at 19:26
• no, just horizontal. Jul 12 '20 at 19:34
• How come it's just horizontal and not vertical as well? Jul 12 '20 at 19:36
• a verticle force would imply the contact force is pushing up on the ball but that's not happening. Jul 12 '20 at 19:42
• Got it. Thanks. Jul 12 '20 at 19:44