How Do You Calculate the Angle of Incline with Friction Involved?

[db7db7]

Homework Statement

On a certain incline plane, a body weighs 50N and requires a force of 35N to pull it up the incline. If the coefficient of friction is 0.25, find the angle of incline. (in degrees)

I've been stuck on this for two days now.

Homework Equations

F=W sin a + u W cos a

The Attempt at a Solution

35N= 50N*sin a + 0.25*50N*cos a

I'm not sure how to isolate for "a".

 

[Delphi51]

I don't know how to isolate "a" either, but you can certainly find the answer to any desired accuracy. You could graph the y = 50N*sin a + 0.25*50N*cos a on your calculator and see where it intersects y = 35. Or put it in a spreadsheet formula and try various values for a until you get an answer of 35.

An interesting approach is to use the method of solving triq equations where you replace cos a with x and sin a with y. This gives a linear equation of the form y = mx + b which you can easily graph. Also graph the unit circle. The intersections of the line and circle are the x,y solutions and the angle counterclockwise from the x-axis is the solution for angle a. You can actually solve this without using the graph if you write x² + y² = 1 for the unit circle and then solve the two equations simultaneously (solve the linear one for y and sub into the circle equation). You end up with a quadratic equation that you can solve to get x and then use x = cos(a) to find the value of a.

 

[kerimek]

Can anyone help?

The angle of incline, also known as the angle of elevation, is the angle between the inclined plane and the horizontal ground. It is typically denoted as "θ" and can be calculated using trigonometric functions.

In this problem, we can use the equation F = Wsinθ + μWcosθ, where F is the force required to pull the body up the incline, W is the weight of the body, and μ is the coefficient of friction.

Substituting the given values, we get 35N = 50Nsinθ + 0.25*50Ncosθ. To isolate for θ, we can use algebraic manipulation and trigonometric identities.

35N - 0.25*50Ncosθ = 50Nsinθ

Divide both sides by 50N and use the identity sinθ/cosθ = tanθ, we get:

35/50 - 0.25*cosθ = tanθ

0.7 - 0.25*cosθ = tanθ

Solving for θ, we get:

θ = arctan(0.7 - 0.25*cosθ)

Using a calculator, we get θ ≈ 31.8 degrees.

Therefore, the angle of incline is approximately 31.8 degrees.