- #1

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In summary, the conversation discusses how to compute the force along the incline ($F_{inc}$) and the force normal ($F_n$) for a 10 kg block sitting on an inclined board at a 30 degree angle with the horizontal. The force of gravity ($F_g$) is decomposed into two orthogonal vectors, with $F_{inc}$ and $F_n$ being the legs of a 30-60-90 right triangle. By using trigonometric functions, it is determined that $F_{inc} = 0.5F_g$ and $F_n = 0.87F_g$.

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- #2

DavidCampen

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

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

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

- #3

WMDhamnekar

MHB

- 378

- 28

Hello,DavidCampen said:You are decomposing $${F}_{g}$$ into 2 orthogonal vectors so

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

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.

- #4

DavidCampen

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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.

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

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- #5

WMDhamnekar

MHB

- 378

- 28

DavidCampen said: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?

- #6

DavidCampen

- 65

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

I edited my post just above to match this.

- #7

WMDhamnekar

MHB

- 378

- 28

DavidCampen said:

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

I edited my post just above to match this.

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

- #8

DavidCampen

- 65

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Yes.Dhamnekar Winod said:So, the $F_{normal}=84.957N$ and $F_{inc}=49N$

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 ${F}_{n} = {F}_{g}sin(60) = 0.87{F}_{g}$.

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The force normal is the perpendicular force exerted by a surface on an object that is in contact with it. The force along the incline is the component of the force acting on an object that is parallel to the incline or ramp.

The force normal and force along the incline are related by the angle of the incline. As the angle of the incline increases, the force along the incline also increases, while the force normal decreases.

The force normal and force along the incline are important in determining an object's motion because they are the two components of the net force acting on the object. The force normal helps to support the weight of the object, while the force along the incline helps to move the object up or down the incline.

The force normal can be calculated using the formula F_{N} = mg cosθ, where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline. The force along the incline can be calculated using the formula F_{parallel} = mg sinθ.

The weight of an object affects the force normal and force along the incline because they are both dependent on the mass of the object. A heavier object will have a greater force normal and force along the incline compared to a lighter object, assuming all other variables remain constant.

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