An 100.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 29.0° to the ground.

(b) The interaction partner of this normal force has what magnitude and direction?
Magnitude 87.46 N
Direction 61° below the horizontal?

c) What is the static frictional force exerted on the crate by the ramp?
48.48 N

(e) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Magnitude 99.99 N
Direction ° above the horizontal?

All I need to know is the answer and explanation for e) direction in degree because I got tan-1 (87.46/48.48) I got 60.99 and 60.99-29= 31.99 degree but its wrong I tried 60.99 and 31.99 both are wrong please someone help me out. Thank you!

Contact force = sqrt(87.46² + 48.48²) = sqrt (7679+2350) = 9999.6 = 10000,

tan φ = N/F(fr) = 87.46/48.48 = 1.8,
φ = 60.99=61º.
Direction is 61º +29º=90º.

To find the direction of the contact force exerted on the crate by the ramp, you can use trigonometry. You're correct in using the inverse tangent (tan^-1) function to find the angle.

First, let's break down the contact force into its two components: the normal force (87.46 N) and the frictional force (48.48 N).

The normal force is perpendicular to the ramp, so it acts vertically upwards.

The frictional force acts parallel to the ramp, opposing the crate's tendency to slide down.

By using the Pythagorean theorem, you can find the magnitude of the contact force:

Contact force = square root((normal force)^2 + (frictional force)^2)
Contact force = square root((87.46 N)^2 + (48.48 N)^2)
Contact force ≈ 99.99 N

Now, let's find the direction of the contact force. We'll use the inverse tangent function:

θ = tan^-1(opposite/adjacent)
θ = tan^-1(frictional force/normal force)
θ = tan^-1(48.48 N/87.46 N)
θ ≈ 29.87°

Therefore, the direction of the contact force is approximately 29.87° above the horizontal.

To find the direction of the contact force exerted on the crate by the ramp, you can use trigonometry. The contact force is the vector sum of the normal force and the frictional force.

First, let's find the x-component of the contact force. Since the normal force and the frictional force are perpendicular components, the x-component of the contact force is equal to the frictional force: Fx = 48.48 N.

Next, let's find the y-component of the contact force. The normal force makes an angle of 61° below the horizontal, so its y-component can be calculated as: Fy = Normal Force * sin(61°)
Fy = 87.46 N * sin(61°)
Fy ≈ 75.80 N

Finally, we can find the magnitude and direction of the contact force using the x and y-components. The magnitude of the contact force (F) is the square root of the sum of the squared x and y-components: F = sqrt(Fx^2 + Fy^2)
F = sqrt((48.48 N)^2 + (75.80 N)^2)
F ≈ 89.86 N

To find the direction of the contact force, you can use the inverse tangent function: θ = tan^(-1)(Fy / Fx)
θ = tan^(-1)(75.80 N / 48.48 N)
θ ≈ 56.11°

Therefore, the magnitude of the contact force is approximately 89.86 N, and the direction is approximately 56.11° above the horizontal.