Force Applied to Box on Incline

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SUMMARY

The discussion focuses on calculating the normal force acting on a box resting on an inclined surface with an angle of θ=20.0°. The box has a mass of m=36kg and an applied vertical force of Fapp=48N. The coefficients of friction are μs=0.240 for static friction and μk=0.150 for kinetic friction. To determine the normal force, one must first assess whether the static friction is overcome by the applied force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to draw and interpret free body diagrams
  • Basic trigonometry for resolving forces on an incline
NEXT STEPS
  • Calculate the normal force using the equation N = mg cos(θ) - Fapp sin(θ)
  • Determine if the static friction is overcome by comparing Fapp with the maximum static friction force
  • Learn about the transition from static to kinetic friction in inclined planes
  • Explore the effects of varying angles on normal force and friction
USEFUL FOR

Physics students, educators, and anyone studying mechanics, particularly those focusing on forces acting on objects on inclined planes.

morgane28v
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A force is applied to a box that is initially at rest on an inclined surface. The incline is at an angle θ=20.0° above horizontal. The mass is m=36kg and the vertical force is Fapp= 48N. Between the box and the inclined surface the coefficient of static friction, μs=.240 and a coefficient of kinetic friction, μk=.150. Use this information to fine the normal force in this situation.

I am confused on where to start. Do I have to use both friction coefficients?
 
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Is the applied force vertically up or vertically down? Either way, draw the free body diagram.
Which coefficient to use depends on whether things move. You know it is initially at rest, so try supposing it stays that way. Use the information provided to find whether static friction is overcome. If it is, switch to using the coefficient for kinetic friction.
 
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?

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