MHB Finding the force on an inclined plane

AI Thread Summary
To find the force required to accelerate a 150 kg crate up a frictionless incline at 30° with an acceleration of 7.1 m/s², the net force must equal the mass times the acceleration. The force due to gravity along the incline is calculated using the sine function, specifically F_g = m * g * sin(30°). The net force equation becomes F_net = m(7.1 + g/2), where g is the acceleration due to gravity (9.8 m/s²). Substituting the values gives a required force of 1800 N. The calculations confirm the accuracy of the solution.
cbarker1
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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
 
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Let's assume the applied force is parallel to the incline plane...and so we require:

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

What is the magnitude of the force due to gravity ($F_f$) along the plane?
 
MarkFL said:
Let's assume the applied force is parallel to the incline plane...and so we require:

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

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

I believe it is $$m\cdot g\cdot \cos\left({30}\right)$$.
 
Cbarker1 said:
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?
 
MarkFL said:
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.
 
Cbarker1 said:
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:

free_body.svg.png


What do you find for $F_{\text{net}}$?
 
MarkFL said:
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.
 
Cbarker1 said:
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:

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

And so we may write:

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

Next, use:

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

So, what do you get?
 
the answer is 1800 N
 
  • #10
Cbarker1 said:
the answer is 1800 N

Yes, I concur. (Yes)
 
  • #11
I did the problem correctly.
 
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