Find Friction Force of Box with Mass of 135 kg at 35° Angle

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The discussion revolves around calculating the friction force acting on a box with a mass of 135 kg being pushed at a 35° angle. The box moves horizontally at a constant velocity, indicating that the net force is zero and the friction force equals the horizontal component of the applied force. The worker applies a force of 2030 N at an angle of 215°, leading to a calculated friction force of approximately 1663 N. Participants confirm that the friction force should match the horizontal force due to the box's constant velocity. The conclusion emphasizes that the friction force is about 1663 N.
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The worker pushes downward and to the left on the box at an angle of 35°. The box has a mass of 135 kg. The box moves horizontally across the floor to the left at a constant velocity. The force applied by the worker is 2030 N∠215°. What is the value of the friction force?Fgravity · sin35° = Fgravityx
Ffriction + Fgravityx = m · a
f = μ · Fssin35° =
Fgravity · sin35° = Fgravityx
m · g · sin35° = Fgravityx
(135 kg)(9.81) sin35° = Fgravityx
759.6 N = Fgravityx

Ffriction + (759.6 N) = (135 kg)(0 m/s2)
Ffriction - 759.6 N = 0 N
Ffriction = 759.6 N

I know that this isn't correct, because my study guide is multiple choice. My options are:
a. 1164 N
b. 1324 N
c. 1663 N
d. 2489 N

I'm really not sure what I'm missing or if I am going the wrong way with finding the value of the friction force. My guess is that c. 1663 N is correct, but that is only a guess. I'd like to be able to prove which answer it is. Thank you for any help!
 
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Your guess is right. Your calculations are off quite a bit. What can you conclude about the friction force since the box is moving to the left at a constant velocity?
 
Probably that it is about equal to the Horizontal force, and that the Net Force is 0
 
Student100 said:
Your guess is right. Your calculations are off quite a bit. What can you conclude about the friction force since the box is moving to the left at a constant velocity?

Probably that it is about equal to the Horizontal force, and that the Net Force is 0
 
So the forces are balanced right? So to calculate the frictional force on the horizontal plane why are you using sin and fg? What should you do instead?
 
DaDoctor said:
Probably that it is about equal to the Horizontal force, and that the Net Force is 0
Exactly, so what is the Ff in the horizontal direction?
 
Student100 said:
Exactly, so what is the Ff in the horizontal direction?
... about 1663 Newtons of force since it is a total of 2030N at a 215 degree angle.
 
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DaDoctor said:
... about 1663 Newtons of force since it is a total of 2030N at a 215 degree angle.
Yep. =)
 
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Student100 said:
Yep. =)
Awesome! Thanks a lot!
 
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