MHB Force normal and force along the incline

[WMDhamnekar]

A 10 kg block sits on a board inclined to an angle of $30^\circ$ with the horizontal.

What are $F_{inc}$ (force along the incline) and the $F_n$ (force normal)?How to compute this?

 

[DavidCampen]

You are decomposing $${F}_{g}$$ into 2 orthogonal vectors so

$${F}_{g} = {F}_{inc} - {F}_{n}$$

 

[WMDhamnekar]

You are decomposing $${F}_{g}$$ into 2 orthogonal vectors so

$${F}_{g} = {F}_{inc} - {F}_{n}$$

Hello,

Here $F_g=-98N$. What is $F_{inc}?$ I think it is $F\langle \cos{60^\circ},\sin{60^\circ}\rangle$ What is $F_{normal}?$

They should be orthogonal to each other means their dot product =0.

 

[DavidCampen]

The 2 component vectors will be at right angles to one another. One will have an angle of 30 degree to ${F}_{g}$ and the other 60 degrees.

So think of a 30-60-90 right triangle with ${F}_{g}$ the hypotenuse and ${F}_{inc}$ and ${F}_{n}$ the legs.

 

[WMDhamnekar]

The 2 component vectors will be at right angles to one another. One will have an angle of 30 degree to $${F}_{g}$$ and the other 60 degrees. So $$asin(30) + bsin(60) = {F}_{g}$$. Which of a and b is $${F}_{inc}$$ and which is $${F}_{n}$$ (and the correct sign) you will have to draw a diagram to figure out.

Hello,

$a\sin{30^\circ}$ is + $F_n$ and $b\sin{60^\circ}$ is $-F_{inc}$. How to compute a and b? Would you explain with a graph?

 

[DavidCampen]

Yes, excuse me.

$a = {F}_{g}sin(30)$ and $b = {F}_{g}sin(60)$

I edited my post just above to match this.

 

[WMDhamnekar]

Yes, excuse me.

$a = {F}_{g}sin(30)$ and $b = {F}_{g}sin(60)$

I edited my post just above to match this.

So, the $F_{normal}=84.957N$ and $F_{inc}=49N$

 

[DavidCampen]

So, the $F_{normal}=84.957N$ and $F_{inc}=49N$

Yes.

I drew a diagram. Placing ${F}_{inc}$ on the horizontal axis as one leg of the triangle, ${F}_{n}$ as the vertical leg and ${F}_{g}$ as the hypotenuse I get an angle of 60 deg. between ${F}_{inc}$ and ${F}_{g}$ so the magnitude of ${F}_{inc}$ = ${F}_{g} cos(60) = 0.5 {F}_{g}$ and ${F}_{n} = {F}_{g}sin(60) = 0.87{F}_{g}$.