Forces and friction, moving a crate

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To find the coefficient of kinetic friction for a 1000-N crate pushed with a 260 N force at a 20° angle below the horizontal, the normal force must be calculated using Newton's first law in the vertical direction. The force of friction is determined by multiplying the normal force by the coefficient of friction. When the force is applied at a 20° angle above the horizontal, the crate's acceleration can be calculated using the same coefficient of friction found earlier. The key is to resolve the applied force into its horizontal and vertical components to accurately determine the normal force and subsequent frictional force. Understanding these forces is crucial for solving the problem effectively.
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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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