Forces and friction, moving a crate

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction for a 1000-N crate being pushed with a force of 260 N at a 20.0° angle below the horizontal. The coefficient of kinetic friction can be determined using the relationship between the force of friction and the normal force. Additionally, when the force is applied at an angle of 20.0° above the horizontal, the acceleration of the crate can be calculated by applying Newton's second law, considering the same coefficient of friction derived from the first scenario.

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A 1000-N crate is being pushed across a level floor at a constant speed by a force of 260 N at an angle of 20.0° below the horizontal.
(a) What is the coefficient of kinetic friction between the crate and the floor?

(b) If the 260 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in the figure, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).


I know that the Force of Friction is equal to the Normal force times the coefficient of friction and that you must find the components of the force, but it seems like I'm missing something.
 
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mg0248 said:
I know that the Force of Friction is equal to the Normal force times the coefficient of friction
Yes, so how would you find the normal force? (Hint: Use Newton 1 in the y direction)
and that you must find the components of the applied[/color] force,
Yes, go for it.
 

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