# A bus, a pendulum and acceleration

• jemerlia

## Homework Statement

A bus is descending a uniform 20 degree slope. It brakes with constant deceleration. A pendulum moves 10 degrees away from the vertical to the downward side. Find the acceleration of the bus.

....|
.../|
... ../.|
.../10|
.../...|.../
...O...|.../
....|/
.../.|
.../...|
.../...|
../...20...|vertical
/_______ |___________horizontal_____

F=ma

## The Attempt at a Solution

I can deduce the x-component of the acceleration:

ax = g tan 10 = 1.73ms^-2

The expected answer is 2.0ms^-2. I can't see how to relate the twenty degree slope to the x component of the acceleration to calculate the total acceleration.

Any help or advice gratefully received...

## Answers and Replies

I can deduce the x-component of the acceleration:

ax = g tan 10 = 1.73ms^-2
Careful: The acceleration is not horizontal.

Do this: Analyze force components parallel and perpendicular to the incline surface.

Thanks - point taken - I reworked the expressions for Fy and Fx in terms of string tension:

sumFy = FTcos 10 - mg cos 20

sumFx=FT sin10 - mg sin 20

m x ax = mg cos 20. tan 10 -mg sin 20

ax =g(cos20.tan10 -sin 20)

N.B. ax is x acceleration with xy co-ordinates of the slope!

Sadly the result is still 1.72ms^-2

Perhaps there is an error with FT cos10 and FT sin 10 in the two sum of forces expressions... and perhaps elsewhere...

Help, advice gratefully received...

Thanks - point taken - I reworked the expressions for Fy and Fx in terms of string tension:

sumFy = FTcos 10 - mg cos 20

sumFx=FT sin10 - mg sin 20
What angle does the string make with respect to the slope?