# Calculating direction of magnitude of a crate on a ramp

1. Oct 2, 2008

### WPCareyDevil

1. The problem statement, all variables and given/known data
An 85.0-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 27.0° to the ground.

(a) What is the normal force exerted on the crate by the ramp?

(b) The interaction partner of this normal force has what magnitude and direction?

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

(d) What is the minimum possible value of the coefficient of static friction?

(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?

2. Relevant equations
F=ma
Fr= (N)(k)
mag= sqrt(x^2+y^2)
dir= Tan^(-1) (y/x)

3. The attempt at a solution

I solved everything except the last part (direction) of e.

a) Normal force: 75.3755N
b) Magnitude (partner of normal): 75.3755N, Direction: 62.999 degrees below the horizontal
c) Static friction force: 38.589N
d) coef of friction .50953
e) magnitude of contact force: 85.004N, ??? degrees above the horizontal

The y component would be the normal force/partner [85cos(27)= 75.7355] while the x component is the friction/partner [85sin(27)=38.599]

(75.7355/38.599) * tan^(-1) = 84.18 degrees. Is this number correct, and I just need to account for it being above the horizontal? Or, am I off?

Thank you SO much for your time!

2. Oct 2, 2008

### Staff: Mentor

Your calculation is off. Redo it. Realize that you are calculating the angle with respect to the incline.

A better way to think of this question: The weight and the contact force are the only forces acting on the crate. What's the net force on the crate? So, what must the contact force be (both magnitude and direction)? You should be able to answer without doing any calculations whatsoever.

3. Oct 2, 2008

### WPCareyDevil

Thats what did it for me. Thank you so much!