- #1

-Dragoon-

- 309

- 7

## Homework Statement

Two packages at UPS star sliding down a 20° ramp. Package A has mass of 5.0 kg and coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom?

## Homework Equations

N/A

## The Attempt at a Solution

Here's my drawing of the diagram in the textbook:

http://imageshack.us/photo/my-images/802/dynamicsdiagram1.jpg/

Here are my free-body diagrams of the forces acting on both boxes:

Free-body diagram for box A: http://imageshack.us/photo/my-images/15/freebodydiagramforblock.jpg/

where n is the normal force, FG is the force due to gravity, F_f is the frictional force, F_BonA is the force box b exerts on A, and F_g_x is the x component of the force due to gravity, as the y component of gravity balances out the normal force.

Free-body diagram for block B:

http://imageshack.us/photo/my-images/256/freebodydiagramforblock.jpg/

Where n is the normal force, FG is the force due to gravity, F_f is the frictional force, F_AonB is the force box A exerts on B, and F_g_x is the x component of the force due to gravity. The friction force and the force block A exerts on B oppose its motion.

Now I begin to work, first starting with Newton's second laws:

[itex]\sum f_{A_{y}} = n - F_{G_{y}} = n - mg\cos\theta[/itex]

[itex]\sum f_{A_{y}} = 0 [/itex] so, [itex]n = mg\cos\theta[/itex]

That was for the y component, now for the x component:

[itex]\sum f_{A_{x}} = F_{BonA} + F_{G_{x}} - F_{k_{A}} = m_{A}a_{A}[/itex]

where F_{k_{A} is the kinetic friction on block A, and m_{A}a_{A}

is the net force acting on block A.

Now I further simplify this:

[itex]\sum f_{A_{x}} = m_{B}g\sin\theta + m_{A}g\sin\theta - \mu_{k}m_{A}g[/itex]

Now I substitute in the numbers and solve for the force:

[itex]\sum f_{A_{x}} = (10kg)(-9.8m/s^{2})(\sin20) + (5.0 kg)(-9.8 m/s^{2})(\sin20) - (0.2)(5.0 kg)(-9.8m/s^{2}) = -40.5 N.[/itex]

Okay, this is where I start to doubt my answer, but I continue to solve for the acceleration of box A:

[itex]a_{A} = \frac{\sum f_{A_{x}}}{m_{A}} = \frac{-40.5 N}{5.0 kg} = -8.1m/s^{2}[/itex]

Now I am even more doubtful, as the acceleration does not seem reasonable for the box given the circumstances.

So I use one of the kinematics equations for constant acceleration to find the time it takes for the box to reach the bottom of the ramp:

[itex]S_{1} = S_{0} + V_{0}\Delta t + \frac{1}{2}a_{A}\Delta t^{2}[/itex]

[itex] 0 = 2.0 m + 0 + \frac{1}{2}a_{A}\Delta t^{2}[/itex]

Rearranging and solving for t:

[itex] t = \sqrt{\frac{-4}{-8.1}} = 0.7[/itex] seconds.

The book's answer was 1.48 seconds. I know the book's answer is correct, but they did their solution in a completely different way. I want to know just exactly what I did wrong, what assumptions or inferences I incorrectly made, and where I can improve next time. Thanks in advance.