Abstract question- solving for acceleration up an inclined plane

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SUMMARY

The discussion centers on calculating the acceleration of a safe being pushed up an inclined plane with a horizontal force, taking into account kinetic friction. The net force equation derived is Fnet = Fcosθ - mgsinθ - μmgcosθ, leading to the acceleration formula a = (Fcosθ/m) - gsinθ - μgcosθ. With specific values provided (Fpush = 4000N, m = 500kg, θ = 20°, μk = 0.20), the expected acceleration is 1.777 m/s². The importance of including the vertical component of the applied force in the free body diagram is emphasized to accurately determine the normal force and, consequently, the acceleration.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of forces on inclined planes
  • Familiarity with kinetic friction concepts
  • Ability to perform trigonometric calculations involving angles
NEXT STEPS
  • Review free body diagram techniques for inclined planes
  • Study the effects of normal force on frictional force calculations
  • Learn about the implications of angle θ on force components
  • Practice problems involving forces and acceleration on slopes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking to enhance their teaching of inclined plane problems.

will5656
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Homework Statement



Burglars are pushing, with a horizontal force Fpush, a safe of mass m and coefficient of kinetic friction μk up a slope of angle θ. What is the safe's acceleration (in abstract terms)?

Homework Equations



as= +/-gsinθ (natural accl down a slope)
friction on a slope= μmgcosθ


The Attempt at a Solution



Fnet= Force up the slope- natural accl down - kinetic friction
Fnet= Fcosθ - mgsinθ - μmgcosθ
ma= Fcosθ - mgsinθ - μmgcosθ
a= Fcosθ/m - gsinθ - μgcosθ

I thought this was right but we have numbers to plug into check for correctness:
fpush= 4000N
m= 500kg
θ= 20°
coeff of kf= 0.20

and acceleration is supposed to be 1.777 m/s2

I can't seem to arrive at this answer.
 
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Hi will5656, Welcome to Physics Forums.

When you draw your free body diagram for the safe, be sure to include the portion of the applied force Fpush which is adding to the normal force...

You've accounted for the "uphill" portion of the force with Fcosθ, but another component of that force is acting to push the safe against the slope. There will be consequences...
 
Remember that the vertical component of the pushing force is increasing the safe's normal force!
 

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