Solve Sliding Box Problem: Force Normal, Constant Velocity & Acceleration

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SUMMARY

The discussion focuses on solving the Sliding Box Problem involving a 25 kg trunk being pulled up a 20-degree incline with a coefficient of kinetic friction of 0.5. The calculations for the normal force yield a value of 230 N using the formula Fn = (m)(g)(cos20). For maintaining constant velocity, the net forces must balance, leading to a corrected equation that includes both the gravitational and frictional forces acting down the ramp. The necessary force to accelerate the trunk at 0.5 m/s² requires careful consideration of all forces acting on the trunk, including friction and gravitational components.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of trigonometry and vector components
  • Familiarity with the concepts of friction and normal force
  • Ability to solve equations involving forces and acceleration
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  • Study the derivation of forces on inclined planes in physics
  • Learn about the effects of friction on motion in different scenarios
  • Explore advanced topics in dynamics, such as non-uniform acceleration
  • Practice solving similar problems involving forces and motion on inclines
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for examples of force analysis on inclined planes.

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Homework Statement


A student is pulling a 25 kg trunk up an incline that has an angle with the horizontal of 20 degrees. She has a strap tied to the trunk which is parallel to the surface of the incline. If the coefficient of kinetic friction between the incline and the trunk is 0.5 fine
a)Force Normal
b) the force necessary to keep the trunk moving at a constant velocity.
c)the force she would have to apply to accelerate the trunk up the incline at 0.5 m/s squared.

Homework Equations


f(net)=ma
coeff.=(force friction)/(force normal)

The Attempt at a Solution



So I think I found (A) to be force normal = 230N by doing Fn=(m)(g)(cos20)
I tried doing this for B (25)(-9.81)(sin20)-(0.5)(25)(-9.81)(cos20) + ForceX =0
I got 31 but that's not right...
 
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(25)(-9.81)(sin20) + (0.5)(25)(-9.81)(cos20) + ForceX = 0

The friction force opposes motion, so it must point down the ramp, and the weight component parallel to the ramp must also point down the ramp. So these two force must have the same sign (direction). Both oppose the pulling force up the ramp.
 

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