# Physicsfinding the force on an inclined plane

#### Cbarker1

##### Active member
Dear Everybody,

What force (in N) must be applied to a 150.0
kg crate on a frictionless plane inclined at 30° to cause an acceleration of 7.1 m/s2 up the plane?

Work:
I know the sum of the force in the x direction must be equal to mass multiply by acceralation.

Thanks
Carter

#### MarkFL

Staff member
Let's assume the applied force is parallel to the incline plane...and so we require:

$$\displaystyle F_{\text{net}}-F_g=m\cdot7.1\,\frac{\text{m}}{\text{s}^2}$$

What is the magnitude of the force due to gravity ($F_f$) along the plane?

#### Cbarker1

##### Active member
Let's assume the applied force is parallel to the incline plane...and so we require:

$$\displaystyle F_{\text{net}}-F_g=m\cdot7.1\,\frac{\text{m}}{\text{s}^2}$$

What is the magnitude of the force due to gravity ($F_f$) along the plane?
I believe it is $$m\cdot g\cdot \cos\left({30}\right)$$.

#### MarkFL

Staff member
I believe it is $$m\cdot g\cdot \cos\left({30}\right)$$.
Think about the cases where the plane is either vertical or horizontal...does the cosine function makes sense?

#### Cbarker1

##### Active member
Think about the cases where the plane is either vertical or horizontal...does the cosine function makes sense?
NO, it does not make any sense. so it must be sine function.

#### MarkFL

Staff member
NO, it does not make any sense. so it must be sine function.
Yes, it is the sine function...here's a free-body diagram:

What do you find for $F_{\text{net}}$?

#### Cbarker1

##### Active member
Yes, it is the sine function...here's a free-body diagram:

What do you find for $F_{\text{net}}$?
Is that $$F_N=mass\cdot 7.1+m\cdot g\cdot sin 30$$ correct?

Sorry, I have misread the question that there is a force pushing it up the inclined plane.

#### MarkFL

Staff member
Is that $$F_N=mass\cdot 7.1+m\cdot g\cdot sin 30$$ correct?

Sorry, I have misread the question that there is a force pushing it up the inclined plane.
That's correct, although we know:

$$\displaystyle \sin\left(30^{\circ}\right)=\frac{1}{2}$$

And so we may write:

$$\displaystyle F_{\text{net}}=m\left(7.1+\frac{g}{2}\right)\text{ N}$$

Next, use:

$$\displaystyle m=150.0\text{ kg},\,g=9.8\,\frac{\text{m}}{\text{s}^2}$$

So, what do you get?