Calculating Forces and Friction in a Moving Package

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The discussion focuses on calculating the forces acting on a package being moved by a car. The speed of the car was determined to be 36.58 m/s, leading to a calculated force of 1503N required to maintain the package's speed. A free body diagram (FBD) was drawn, resulting in the equation mu*Nsin60 + Nsin30 = mg, with mg calculated as 1323. The coefficient of friction (mu) was estimated to be 0.3. The thread seeks clarification on the complete equations for forces in both horizontal and vertical directions.
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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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