How Does Tension at an Angle Affect Friction and Motion?

[giacomh]

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

 

[netgypsy]

I didn't get the same answer and you have left out steps so let's start from the beginning summing the forces in the x and y direction

I think it comes out to be the same equation you have so it may just be that I ran it all at once rather than in two parts.

Fx max = Ff = mu Fn

Fy +Fn = Fw

If you solve equation 2 for Fn and put in your sin and cos to get the components you can then solve the first equation for F.

Run it one more time and check your answer as I didn't get quite the same thing. It could have been rounding but just double check. I don't see any errors in part a.

for part B how did you get the value of 19.6 for the x component of the force which is 20.5 (.866)? I get 17.7 which changes the answer.For c I get the same thing.

 

[giacomh]

Thank you for your response!

I got a slightly different answer for part a after changing some things. Here is my work:

Fx=uN
N=mg-Fsin30
Fx=Fcos30

Fcos30=u(mg-Fsin30)
Fcos30=umg-uFsin30
F(cos30-usin30)=umg
F=umg/(cos30-usin30)
F=(.2*10*9.8)/(cos30-.2sin30)

F=25.58N

Like I said, finding the Fn at an angle has been tripping me up.

 

[netgypsy]

I think you have a sign error

Isn't it F cos 30 = mumg - muFsin30
so it would be F(cos30 + mu sin30) = mu mg

You did it correctly the first time. I just got 20.3 rather than 20.5

If you always draw your forces and add the ups and set them equal to the sum of the downs you won't usually have a problem as long as they are in equilibrium.

 

[asteriatic]

m/s^2

First of all, great job on attempting to solve this problem! It's important to practice and verify your answers before a test. Let's go through your solution step by step to make sure everything is correct.

a. Your equation for the maximum static friction force is correct: Fsmax=μsN. However, in your calculation, you used the weight of the box (mg) instead of the normal force (N). Remember, the normal force is equal to the weight of the object only if it is sitting on a horizontal surface. In this case, the normal force will be less than the weight of the box because the cord is pulling up at an angle. So the correct equation should be: Fsmax=μs(Ncos30), where Ncos30 is the component of the normal force in the horizontal direction. Then, you can solve for the pulling force F: F=Fsmax/(cos30) = 17.82N.

b. Your setup for the free body diagram is correct, but your calculation for the acceleration is not. Remember, the net force in the x-direction is equal to the pulling force minus the kinetic friction force: ΣFx= F-μkN. Then, using Newton's second law, ΣFx= ma, we can solve for the acceleration: a=(F-μkN)/m. Plugging in the values, we get a=(19.6-0.1(98-20.57sin30))/10 = 1.083 m/s^2.

c. In this case, you are correct that the normal force is equal to 0, since the box is about to lift off the floor. However, the pulling force will also be equal to the maximum static friction force, since the box is just about to move. So the equation should be: F=μsNcos30. Plugging in the values, we get F=0.2(98cos30) = 16.9 N. Then using Newton's second law, ΣFx=F=ma, we can solve for the acceleration: a=F/m = 16.9/10 = 1.69 m/s^2. This is the acceleration just before the box is lifted off the floor.

Overall, you did a great job with your calculations and understanding the concepts of friction and tension at an angle. Just remember to double check your equations and units to make