Work with friction

1. May 7, 2010

dator

1. The problem statement, all variables and given/known data
In many neighbourhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50kg and the adult does 2.2x10^3 J of work pulling the two 60m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.

a) Draw a FBD of the wagon
b) Determine the magnitude of the force applied by the parent.
c) Determine the angle at which the parent is applying this force.

2. Relevant equations
F_friction = mu*F_normal
a^2=b^2+c^2
Work=Force*(delta)d*cos(theta)

3. The attempt at a solution
Just want to make sure I did this right. And do I have to include the units when i write formulas out? (N*M) etc. It it important too or it doesn't matter? Thanks in advance!

Variables:
Work = 2.2x10^3 J
(delta)d= 60m
mu = 0.26
Force = ?
Theta =?

In Terms of y:
F_normal=-(mg+Force*sin(theta))

In Terms of x:
F_friction = mu*F_normal

Since speed is constant acceleration=0, therefore F_x+F_friction = 0 or (F_x=-F_friction)

Let:
Force*sin(theta) = F_y
Force*cos(theta) = F_x

In terms of x:

Work = Force*(delta)d*cos(theta)
Force*cos(theta) = Work/(delta)d
F_x = 2.2x10^3/60
F_x = 36.77777776

Therefore F_x+F_friction = 0

F_x+mu*F_normal = 0

Sub in numbers.

36.7+(-1)(0.26)[(50)(-9.8)+Force*sin(theta)= 0
36.7+(-0.26)[-490+Force*sin(theta)]= 0
36.7+127.4+(-0.26)(Force*sin(theta) = 0
Force*sin(theta) = -164.1/-0.26
Force*sin(theta) = 631.1538462
F_y = 631.1538462

F_y = 631.1538462
F_x = 36.7

Force = sqrt(F_y^2+F_x^2)
Force = 632.38N

opp = F_y

(theta) = tan^-1(F_y/F_x)
(theta) = 86.6
(theta) = 87

a)cant draw it here, but the F_normal has the F_y going parallel with it, and F_fr is opposing the F_x.

b) 632.38N
c)(theta) = 87

2. May 7, 2010

collinsmark

Hello dator,

Welcome to Physics Forums!

According to my calculations, something is not right with the very last line above.

I think the root of the error might come from your equation:
The vectors are not adding up (or subtracting up) right. Rearrange the above equation to put all the terms on one side of the equation; then set all that to zero. Compare that to your FBD.

In the future, if might be beneficial to work the problem in smaller steps, rather than trying to combine everything into one big monster step. For example, after you calculate the force of friction (which you did early, and that's okay), calculate the normal force. That way you can check if things make sense before moving on to the next step.

3. May 7, 2010

dator

Thank you! :)

Bah, I should plugged the numbers back in and checked the statement I made, F_x+F_friction=0. The F_y I solved for does not give me a true statement.

Statements should be as follows then,
F_normal = -(m*-g+Force*sin(theta))
F_normal = (mg-Force*sin(theta))

Therefore, F_x+F_friction = 0 is:

Force*cos(theta)+(mu)(mg-Force*sin(theta)) = 0
36.7+(0.26)(490-Force*sin(theta)) = 0
36.7+127.4-(0.26)(Force*sin(theta)) = 0
Force*sin(theta) = -164.1/-0.26
Force*sin(theta) = 631.1538462

Double check....

36.7+(0.26)(490-631.1538462)=-0.0000001
So basically zero, blame it to roundoff.

Does that look good now?

4. May 9, 2010