Calculating Forces and Friction in a Moving Package

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SUMMARY

The discussion centers on calculating the forces acting on a package being moved by a car traveling at a speed of 36.58 m/s. The force required to maintain the package's speed is determined to be 1503 N using the formula F = (mv^2)/r. Additionally, the coefficient of friction (μ) is calculated to be 0.3 based on the equation μ*Nsin60 + Nsin30 = mg, where mg equals 1323 N. The user seeks clarification on the complete equations for both vertical and horizontal forces acting on the package.

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  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams (FBD)
  • Knowledge of friction coefficients and their calculations
  • Proficiency in applying circular motion formulas, specifically F = (mv^2)/r
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guoruiwu
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For this question, first I tried finding the speed at which the car is moving at (36.58m/s), but I'm not sure if that's entirely relevant to solving the question.

The only way the package is moving is from the force applied from the car, so using the formula F = (mv^2)/r, I found that the force to keep the package the same speed as the car is 1503N.

Then I drew a FBD for the package, and found that in the vertical direction, mu*Nsin60+Nsin30 = mg = 1323. So, therefore mu has to be c (0.3).

I'm not sure if this is right or not though, any help would be appreciated.
 

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guoruiwu said:
The only way the package is moving is from the force applied from the car, so using the formula F = (mv^2)/r, I found that the force to keep the package the same speed as the car is 1503N.
That gives you the net force, but what's the full equation for forces in the horizontal direction?
Then I drew a FBD for the package, and found that in the vertical direction, mu*Nsin60+Nsin30 = mg = 1323.
What's the corresponding equation for the horizontal direction?
 

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