(adsbygoogle = window.adsbygoogle || []).push({}); 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

tan(theta) = opp/adj

opp = F_y

adj = F_x

(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

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# Homework Help: Work with friction

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