# Homework Help: Suitcase Dynamics Problem

1. Jan 31, 2010

### temaire

1. The problem statement, all variables and given/known data
At the airport, you pull a 18-kg suitcase across the floor with a strap that is at an angle of 35° above the horizontal. Find (a) the normal force and (b) the tension in the strap, given that the suitcase moves with constant speed and that the coefficient of kinetic friction between the suitcase and the floor is 0.38.

2. Relevant equations
F = ma
Fn = -mg
Fk = (N)(u)

3. The attempt at a solution

a.)
Fn = normal force
u=coefficient of friction
t=tension
Fx= x-component or horizontal force
to get the normal force,
Fn=Fx/u
Fx is not given so you have to get t first to know Fx

b.)
W=weight of suitcase
Fx= x-component or horizontal force = t(cos 35)
Fy=y-component = t(sin 35)
Fn=W-Fy
you can also use the first formula which is Fn=Fx/u
since the two formulas are both solving for Fn, you can equate them to each other to get t first. Fx and Fy both have t so substitute both
W-Fy=Fx/u
W-[t(sin 35)] = t(cos35)/u
get W first
W=mg
W=(18kg)(9.8m/s^2)
9.8m/s^2 is the constant for acceleration due to gravity. We have weight because of gravity right?
W=176.4 N or Newtons
Now you can solve for t
Again,
W-[t(sin 35)] = t(cos35)/u
176.4 N - [t(sin 35)] =(t cos 35)/0.38
0.38 * {176.4 N - [t(sin 35)]} =t(cos35)
67.032 N - t(0.217959045) =t (cos 35)
- t(0.217959045) - t (cos 35) = -67.032 N
-1.037111089 (t) = -67.032N
t=64.63338471 N -------------- answer for b

Continuation of a:
Now that you already have tension or t you can get Fn
get Fx first:
Fx=t(cos35)
Fx=64.63338471 N (cos 35)
Fx=52.94 N
Fn=Fx/u
Fn=52.94 N/0.38