Physics Homework Help: Solving for Coefficient of Friction & Acceleration

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SUMMARY

The discussion focuses on solving a physics homework problem involving a 95.0 kg crate on a truck's tilted bed. The coefficient of static friction is calculated as 0.445 when the bed is tilted at 24.0°. When tilted at 30.0°, the acceleration of the crate is determined to be 2.354 m/s², using the coefficient of kinetic friction of 0.35. To find the final velocity after sliding 1.5 m, participants are advised to apply kinematic equations, specifically utilizing the known acceleration and displacement.

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  • Understanding of static and kinetic friction concepts
  • Familiarity with free body diagrams
  • Knowledge of Newton's second law of motion
  • Basic proficiency in kinematic equations
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  • Study the derivation of friction coefficients in physics
  • Learn how to construct and analyze free body diagrams
  • Explore Newton's laws of motion in greater detail
  • Practice solving kinematic equations for various motion scenarios
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Students studying physics, particularly those tackling problems related to friction, motion on inclined planes, and kinematic equations.

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Homework Statement



a) a truck tilts its flat bed slowly to dispose of a 95.0 kg crate. For small angles the crate stays put but just begins to slide when the angle of tilt is 24.0°. Make a free body diagram for the crate when it is on the verge of sliding, clearly showing the relevant forces and calculate the coefficient of static friction between the crate and the truck’s bed.

b)If the bed is tilted at 30.0°, what will be the acceleration of the crate if coefficient of kinetic friction is 0.35?

c)If the crate slides 1.5 m from rest, what is its velocity at the bottom for case b)?

Homework Equations





The Attempt at a Solution


a)The coefficient friction at the point where the crate just starts sliding = tan (the angle) = tan 24 = .445

b)The force of gravity perpendicular to the plane is called the normal force and it is equal to mg cos(angle) = (95)(9,8)cos(3) = 806.246
the force of gravity parallel to the plane is
mg sin(angle) = (95)(9.8)sin(3) = 465.5
force of friction parrellel to plane(Ff) = uFn = (.3)(806.246) = 241.872
the net force = Fn – Ff = 4.65 – 241.872 = 223.629
divide this by the original mass for acceleration = 223.629/95 = 2.354

c) i don't know how to approach this problem...
 
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terryW16 said:

Homework Statement




c)If the crate slides 1.5 m from rest, what is its velocity at the bottom for case b)?

i don't know how to approach this problem...
You have the acceleration and displacement down the incline...use one of the kinematic equations to solve for the velocity at the bottom.
 

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