Calculating direction of magnitude of a crate on a ramp

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SUMMARY

The discussion focuses on calculating the forces acting on an 85.0-N crate resting on a ramp inclined at 27.0°. The normal force exerted on the crate is 75.3755 N, while the static frictional force is 38.589 N, leading to a coefficient of static friction of 0.50953. The magnitude of the contact force is determined to be 85.004 N, with the direction needing clarification. The correct approach emphasizes understanding the net forces acting on the crate rather than relying solely on calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its coefficient
  • Familiarity with trigonometric functions in physics
  • Ability to resolve forces into components
NEXT STEPS
  • Study the concept of net force and equilibrium in physics
  • Learn about vector resolution and force diagrams
  • Explore static friction and its applications in real-world scenarios
  • Review trigonometric identities and their use in force calculations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding force calculations on inclined planes and static friction dynamics.

WPCareyDevil
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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!
 
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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!
 

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