15) In Question 14, if the ramp makes an angle of 20 degrees with the level ground. Find the magnitude of the force tending to lift the crate vertically.

++++Textbook Answer for Question 15: 108.3 N

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14) A crate is being dragged up a ramp by a 125 N force applies at an angle of 40 degrees to the ramp. Find the magnitude of the force in the direction of motion (Answer is 95.8 N).

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+++++Please provide a step by step solution to this problem (question 15)! I really don't know what I'm doing! I know the answer, but I don't know how to get that answer! Thank you.

14)

F vector at 40 degrees to motion direction Call Unit vector up ramp parallel to ramp Au.
The component of F in the direction of Au is
F dot Au = |F| * 1 cos (40) = 125(.766) = 95.8 Newtons

15) The 125 N is at 20+40 = 60 degrees from horizonal, so at 30 degrees from vertical
125 cos 30 = 108.3 N

To find the magnitude of the force tending to lift the crate vertically, we can break down the given information step by step and use trigonometry.

Step 1: Understand the problem.
- A crate is being dragged up a ramp by a 125 N force.
- The force is applied at an angle of 40 degrees to the ramp.
- We want to find the magnitude of the force tending to lift the crate vertically.

Step 2: Analyze the forces.
- The 125 N force can be resolved into two components: one parallel to the ramp and one perpendicular to the ramp.
- The force parallel to the ramp is the force in the direction of motion (Question 14) and has a magnitude of 95.8 N.
- We need to find the force perpendicular to the ramp, which represents the force tending to lift the crate vertically (Question 15).

Step 3: Apply trigonometry.
- Since we know the angle between the applied force and the ramp is 40 degrees, we can use trigonometric functions.
- The force parallel to the ramp is the adjacent side of a right triangle, and the force perpendicular to the ramp is the opposite side.
- To find the force perpendicular to the ramp, we can use the sine function.
- The sine of an angle is equal to the ratio of the length of the side opposite the angle to the length of the hypotenuse.
- In this case, the hypotenuse is the 125 N force applied at an angle of 40 degrees.
- Therefore, the force perpendicular to the ramp can be calculated as follows:

Force perpendicular = Force applied * sine(angle)
= 125 N * sin(40 degrees)
≈ 125 N * 0.6428
≈ 80.35 N

So, the magnitude of the force tending to lift the crate vertically is approximately 80.35 N.