How Much Force is Needed to Move a Crate with Friction?

[elsternj]

Homework Statement

A factory worker pushes a 28.0kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 31 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26. What magnitude of force must the worker apply to move the crate at constant velocity?

Homework Equations

F=ma
f=mu*N

The Attempt at a Solution

F=ma
w=274.4
N = 274.4
it is saying to move the crate at a constant velocity which would make a=0.
So F_x=Fcos31-f=ma
F_x=Fcos31-(.26)(274.4)=m(0)
F_x=Fcos31-71.3=0
F_x=Fcos31=71.3
F_x=F=83.2 which is not the right answer.

I am doing something wrong here. Can't quite figure out what it is.


This is ultimately a work problem but I will be fine once I can get this part. I'm lost somewhere and just need to be steered on the right track.

 

[supratim1]

the net normal force will be sum of mg and the component of force along normal. so can you tell what would be N then?

 

[elsternj]

The y component of force is pushing down on the crate so N=mg+Fsin31, right? But I still don't know what F is.

 

[elsternj]

This would make my sum of Forces in the x direction as:
Fcos31-(.26)(274.4+Fsin31)=0
am I right in setting this equal to 0?

 

[SammyS]

This would make my sum of Forces in the x direction as:
Fcos31-(.26)(274.4+Fsin31)=0
am I right in setting this equal to 0?

Looks good to me.

 

[elsternj]

okay i am brushing up on a lot of math as I am in this course. So this is my math step by step. It gave me close to the right answer so I don't know if the mistake is in my math or the online homework system rounding funny which it is known to do.

Fcos(-31)-(.26)(274.4+Fsin(-31))=0
Fcos(-31)-71.34+Fsin(-31)(.26)=0 (distributed the .26)
Fcos(-31)-71.34 - .13F = 0 (multiplied .26 with sin(-31))
Fcos(-31) - .13F = 71.34
(F)(cos(-31)-.13) = 71.34 (Factored out my F)
(F)(.73) = 71.34
F = 97.72

The correct answer is 99. I just want to make sure that I am at least doing my math right.

 

[SammyS]

Solve for F before plugging in numbers. If that's too much of a problem, then don't round of intermediate so much.

(.26)sin(-31°) = -0.13390989...

When you round that to -0.13, that gives a 3% error. Other rounding may make the % error larger.