A bus, a pendulum and acceleration

1. The problem statement, all variables and given/known data

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_____





2. Relevant equations

F=ma

3. 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...
 

Doc Al

Mentor
44,642
966
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...
 

Doc Al

Mentor
44,642
966
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?
 

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