- #1
giacomh
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So I tried working this problem out. I'm fairly certain it will be on my final tomorrow morning, but I can't find anywhere to verify my answers. So if someone can please just check over my math for me, it would help immensely! I have been confused on the normal force when tension at an angle is involved, so that's why I'm so desperate to have this verified. Thank you in advance!
A 10 kg box is sitting on a rough floor. A worker has attached a cord to the box so that he can pull at an angle of 30 degrees above the horizontal. If μk=.1 and μs=.2
a. How hard must the worker pull if the box is to start moving?
F=uN
F= .2 * (mg-Fsin30)
Fsmax= 17.82 N
Fcos30=17.82
F=20.57 N
b.If he continues to pull with the force you found in part a, what will the acceleration be?
Fy=0
Fx= Pull-μkN
Fx-(μK*(mg-Fsin30))=ma
19.6-(.1*(98-20.57sin30))=10a
a=1.083
c. He slowly increases the magnitude of the pulling force. What is the value of the acceleration along the floor just before the box is lifted off the floor?
I know the normal force=0
Fy=mg
Fy=98N
tan30=98/Fx
Fx=169.74N
Fx=max
169.74=10a
a=16.9
A 10 kg box is sitting on a rough floor. A worker has attached a cord to the box so that he can pull at an angle of 30 degrees above the horizontal. If μk=.1 and μs=.2
a. How hard must the worker pull if the box is to start moving?
F=uN
F= .2 * (mg-Fsin30)
Fsmax= 17.82 N
Fcos30=17.82
F=20.57 N
b.If he continues to pull with the force you found in part a, what will the acceleration be?
Fy=0
Fx= Pull-μkN
Fx-(μK*(mg-Fsin30))=ma
19.6-(.1*(98-20.57sin30))=10a
a=1.083
c. He slowly increases the magnitude of the pulling force. What is the value of the acceleration along the floor just before the box is lifted off the floor?
I know the normal force=0
Fy=mg
Fy=98N
tan30=98/Fx
Fx=169.74N
Fx=max
169.74=10a
a=16.9