Calculating direction of magnitude of a crate on a ramp

Click For Summary
An 85.0-N crate rests on a ramp inclined at 27.0°, prompting calculations of forces acting on it. The normal force exerted by the ramp is calculated to be 75.3755 N, with its interaction partner also at this magnitude but directed 62.999 degrees below the horizontal. The static frictional force is found to be 38.589 N, and the coefficient of static friction is approximately 0.50953. The magnitude of the contact force is determined to be 85.004 N, but there is confusion regarding its direction above the horizontal. Clarification suggests that understanding the net forces acting on the crate can simplify determining the correct angle without extensive calculations.
WPCareyDevil
Messages
30
Reaction score
0

Homework Statement


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?

Homework Equations


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

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 horizontalThe 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!
 
Physics news on Phys.org
WPCareyDevil said:
(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?
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. :wink:
 
Doc Al said:
What's the net force on the crate? :wink:

Thats what did it for me. Thank you so much!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
10K
Replies
4
Views
4K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
10K