Inclined Plane Problem

Question

A 230-newton box rests on a plank 5.00 meters long. One end is 1.20 meters higher than the other end. Find the components of the force the box exerts:

a). Parallel to the inclined plane

b). Perpendicular to the inclined plane

Answer 1

First, we need to find the incline of the plane

1. Draw the plank in the form of a right triangle with the ground
2. The plank itself is the diagonal and is 5.00 m long and the height of the triangle is 1.20 m
3. Use trigonometry to find the angle
4. Here, we know the hypotenuse, and the side opposite the incline angle
5. sin(Θ) = opp/hyp
6. sin(Θ) = 1.20m/5.00m
7. Now that we know the angle, we need to look at the force that the box is exerting.
8. The box can only exert weight on the plane, which is given as 230 N, and acts straight down into the Earth.
9. As with any vector, we can break that 230 N down into components. With inclined planes, it's always these parallel and perpendicular ones.
10. Draw your box on the plane and draw your weight vector straight drown and label it 230 N.
11. This will be the resultant vector, so it will be the hypotenuse of a right triangle that you need to draw.
12. From the top of your weight vector, draw a dashed line diagonally down perpendicularly to the plane (representing the "perpendicular component")
13. From the end of your weight vector, draw a dashed line diagonally upwards parallel to the incline of the plane, until it intersects your other dashed line. These should form a right angle. This new line is your "parallel component".
14. Now we need to find the lengths of those 2 dashed lines, with more trigonometry
15. Through some geometry, you can see that the angle between your perpendicular component and the downward weight vector is the same as the angle of your incline.
16. Knowing the hypotenuse of your triangle (230 N), and the angle of your incline, you can now solve for the components using your sin and cos formulas
17. It might be worth writing these down, as these always work for incline planes and they come up quite often:
18. F_parallel = mg sin(Θ)
19. F_{perpendicular} = mg cos(Θ)
20. For your problem, remember that the Weight is already given as 230 N, and W = mg, so you just have:
21. F_parallel = 230 sin(Θ)
22. F_{perpendicular} = 230 cos(Θ)

Check your work:

1. Both answers should be less than 230 N.
2. You can use the Pythagorean Theorem to check:
3. F_parallel² + F_{perpendicular}² = 230²