Finding the force on an inclined plane

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SUMMARY

The discussion focuses on calculating the force required to accelerate a 150.0 kg crate up a frictionless inclined plane at an angle of 30° with an acceleration of 7.1 m/s². The net force equation is established as F_net = m * a, where the gravitational force component along the incline is determined using the sine function. The final calculation reveals that the required applied force is 1800 N, confirming the accuracy of the calculations presented.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of trigonometric functions, specifically sine and cosine
  • Familiarity with free-body diagrams
  • Basic principles of inclined plane physics
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  • Study the application of Newton's laws in various physical scenarios
  • Learn about free-body diagram techniques for analyzing forces
  • Explore the effects of friction on inclined planes
  • Investigate advanced topics in dynamics, such as non-inertial reference frames
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Students and educators in physics, engineers working on mechanical systems, and anyone interested in understanding forces on inclined planes.

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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