Calculating direction of magnitude of a crate on a ramp

In summary, the normal force exerted on the crate by the ramp is 75.3755N. Its interaction partner has a magnitude of 75.3755N and a direction of 62.999 degrees below the horizontal. The static frictional force exerted on the crate by the ramp is 38.589N. The minimum possible value of the coefficient of static friction is 0.50953. The magnitude of the contact force is 85.004N, and its direction is perpendicular to the incline of the ramp.
  • #1
WPCareyDevil
30
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!
 
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  • #2
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:
 
  • #3
Doc Al said:
What's the net force on the crate? :wink:

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

FAQ: Calculating direction of magnitude of a crate on a ramp

What is the formula for calculating the direction of magnitude of a crate on a ramp?

The formula for calculating the direction of magnitude of a crate on a ramp is F = mg(sinθ ± μcosθ), where F is the force, m is the mass of the crate, g is the acceleration due to gravity, θ is the angle of the ramp, and μ is the coefficient of friction.

How do you calculate the angle of the ramp?

The angle of the ramp can be calculated using the equation θ = tan^-1(height/length), where θ is the angle, height is the vertical distance, and length is the horizontal distance.

What is the significance of the coefficient of friction in this calculation?

The coefficient of friction is a measure of the resistance between two surfaces in contact. In this calculation, it is used to determine the amount of friction between the crate and the ramp, which affects the direction and magnitude of the force needed to move the crate.

What units should be used for the variables in the formula?

The mass (m) should be in kilograms (kg), the acceleration due to gravity (g) should be in meters per second squared (m/s^2), the angle (θ) should be in degrees (°), and the coefficient of friction (μ) is unitless.

Can this formula be applied to crates of different masses and on different types of ramps?

Yes, this formula can be applied to crates of different masses and on different types of ramps as long as the variables are adjusted accordingly. The mass and angle of the ramp will affect the force needed to move the crate, while the coefficient of friction will depend on the surfaces in contact.

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