# Acceleration of two packages on an incline

Zontar

## Homework Statement

Two packages at UPS start sliding down a 20 degree incline. Package A has a mass of 5kg and a coefficient of friction of 0.20. Package B has a mass of 10kg and a coefficient of 0.15. Package A is in front of Package B according to a diagram given. The distance between Package A and the bottom of the ramp is 2m. How long does it take for Package A to reach the bottom?

## Homework Equations

Kinematics Equations and Free-Body Diagrams yielded the following breakdown of all the forces, given in this form:

(Force), (x hat) +/- (y hat)
uk_(box) is the kinetic friction coefficient.
Ff is the friction force
Fg/Fn are obvious
X_Y is the force X on Y

## The Attempt at a Solution

The very first thing I did was make a table of forces symbolically:

For package A, assume a tilted axis of 20 degrees, with +x in the direction of the packages' motion.

Fn, 0 + Fn
Fg, mg sin 20 - mg cos 20
Ff, -uk_A(Fn) + 0
B_A, B_A + 0
Fnet, (M_a)(A_a) + 0

For package B, assume an identical axis.

Fn, 0 + Fn
Fg, mg sin 20 - mg cos 20
Ff, -uk_B(Fn) + 0
A_B, -A_B + 0
Fnet, (M_b)(A_b) + 0

The acceleation is constrained by A_a = A_b, allowing us to use one acceleration "a".

I then get these final equations for:
Package A X: mg sin 20 - uk_a(Fn) + B_A = M_a(a)
Package A Y: Fn = mg cos 20
Package B X: mg sin 20 - uk_b(Fn) - A_B = M_b(a)
Package B Y: Fn = mg cos 20

Now, I haven't the foggiest idea what to do. All I see are endless streams of unsolvable equations with two unknowns.

Last edited:

## Answers and Replies

Homework Helper
Each package has a force down the ramp and an opposing friction force.
Write that the sum of the four forces is ma, where m = 15. You should be able to find the acceleration in a jiffy.

Zontar
Each package has a force down the ramp and an opposing friction force.
Write that the sum of the four forces is ma, where m = 15. You should be able to find the acceleration in a jiffy.

That's exactly what I did:

Package A X: mg sin 20 - uk_a(Fn) + B_A = M_a(a)
Package A Y: Fn = mg cos 20
Package B X: mg sin 20 - uk_b(Fn) - A_B = M_b(a)
Package B Y: Fn = mg cos 20

What I didn't understand was how they tied together. The solutions manual says to add the two equations, treating both forces as one big force; however, I don't understand how exactly they did that.

Homework Helper
F = ma
5g*sin(20) - .2*5g*cos(20) + 10g*sin(20) - .15*10g*cos(20) = 15a