How to Calculate the Inclination Angle of a Plane from Acceleration?

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To calculate the inclination angle of a plane from the acceleration of a crate sliding down, one can use the relationship between acceleration and gravitational force components. The acceleration of 7.42 m/s² corresponds to the parallel component of gravity acting down the slope. By applying trigonometric functions, specifically using sine or cosine, the angle can be determined without needing the mass of the crate. The calculated angle is approximately 49.3 degrees, which differs from the book's stated angle of 41 degrees. This discrepancy suggests a possible error in the book's solution.
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Question:
A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of 5.832 m/s after sliding a distance of 2.29m, what is the angle of inclination of the plane with respect to the horizontal?


I first set up a free body diagram then I discovered that finding the acceleration in the x direction would be the best idea.

I have the question down to a=7.42 m/s^2 which is correct for the acceleration, but I am getting stuck on where to go next. The formula I used is:

v^2= Vi^2+2a(x-xi) and that gave me the 7.42 m/s^2. The book says the angle is 41 degrees, but I don't know how to take the acceleration and determine the angle just from the acceleration. Anyone have some pointers?
 
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Think: What forces act on the crate? What force accelerates the crate down the slope?

ehild
 
Thenacceleration along the slope is a component of g acting at an angle... that should help. I reccomend drawing a nice big diagram of the situation with forces on it and looking for a nice right angled trianlge...
 
I did this but the problem never gave a mass. The angle is my desired result. Gravity is acting on the y direction F=mg. So I found acceleration in the x direction so...can I find the angle with the acceleration known even if the mass is not given?
 
You have 2 sides to a triangle - g of 9.81 in a vertical sense, and the component along the slope you calculated earlier. Could some trig be applied here?
 
tarheels88 said:
I did this but the problem never gave a mass. The angle is my desired result. Gravity is acting on the y direction F=mg. So I found acceleration in the x direction so...can I find the angle with the acceleration known even if the mass is not given?

Of course you can. You got a, the acceleration in the x direction. That corresponds to a force F=ma along the slope, red arrow in the picture, and that force is the parallel component of gravity (G, in blue). The arrows make a right triangle (yellow) with an angle α, equal to the inclination angle of the slope. You need only remember the definition of sine, cosine, tangent...

ehild
 

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tarheels88 said:
The book says the angle is 41 degrees

I get the angle as 49.3 degrees to 3 s.f. Correct?
 
Possibly you have found the wrong the angle in the triangle there...
 
lfcprg said:
I get the angle as 49.3 degrees to 3 s.f. Correct?

It has to be correct. The acceleration along the slope due to gravity is gsin(α). The acceleration is 7.426 ms-2, that corresponds to 49.2°, instead of 41.

ehild
 
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