Basic Force: Why to take force of friction from force applied y?

AI Thread Summary
The discussion focuses on understanding why the force of friction is derived from the vertical (y) component of the applied force rather than the horizontal (x) component. It emphasizes that the normal force, which affects friction, is influenced by both the weight of the box and the vertical component of the applied force. The calculation for the normal force combines the weight of the box with the vertical force component, leading to a frictional force that opposes motion. The frictional force is determined using the coefficient of friction and the normal force. This highlights the relationship between the direction of forces and the resulting frictional effects in physics problems.
Arooj
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Homework Statement


A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35^degrees below the horizontal.

a) If uk between the box and the floor is 0.57, how long does it take to move the box 4.00 m, starting from rest.

b) If uk between the box and the floor is 0.75, how long does it take to move the box 4.00 m, starting from rest?

Homework Equations


Just for this problem: Ff = (FsinΘ+W)µ (F=485)
Fnet = FcosΘ - Ff

The Attempt at a Solution


I understand how to do this problem, but why is it that you take the force of friction from the y component of the force rather than the x component of the force?
 
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Arooj said:

Homework Statement


A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35^degrees below the horizontal.

a) If uk between the box and the floor is 0.57, how long does it take to move the box 4.00 m, starting from rest.

b) If uk between the box and the floor is 0.75, how long does it take to move the box 4.00 m, starting from rest?

Homework Equations


Just for this problem: Ff = (FsinΘ+W)µ (F=485)
Fnet = FcosΘ - Ff

The Attempt at a Solution


I understand how to do this problem, but why is it that you take the force of friction from the y component of the force rather than the x component of the force?

Ff=Nμ
Find the value of N.
 
N = FsinΘ + W
N = 485(sin35) + 319
N = 597.2
 
The friction is depend on normal force which is in y direction.
The frictional force is perpendicular to normal force and it opposes the motion.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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