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Physics finding the force on an inclined plane

Cbarker1

Active member
Jan 8, 2013
255
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 8, 2013
255
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 8, 2013
255
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 8, 2013
255
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

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

Cbarker1

Active member
Jan 8, 2013
255
the answer is 1800 N
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

Cbarker1

Active member
Jan 8, 2013
255
I did the problem correctly.