# Rough Inclined Plane SKH Tsang Shiu Tim Secondary School

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```					Motion on a Rough
Inclined Plane
A trolley of mass 0.5 kg is released from rest at a point O on a
rough runway inclined at 30o to the horizontal. It slides down the
runway and the velocity-time graph is given below.
Assume g = 10 m s-2

Determine the friction and normal
V / m s-1         reaction on the trolley.
h
roug
1.6
lley
tro            O
1.2

0.8

0.4

t/s
0      0.2    0.4
3o                        horizontal
A trolley of mass 0.5 kg is released from rest at a point O on a
rough runway inclined at 30o to the horizontal. It slides down the
runway and the velocity-time graph is given below.
Assume g = 10 m s-2
From the velocity-time graph the acceleration of
the trolley is 4 m s-2. The acceleration vector
V / m s-1               points down-slope.
h
roug
1.6
lley
tro           O
1.2

0.8

0.4

t/s
0      0.2   0.4
3o                        horizontal
In the direction perpendicular to the inclined plane, there is NO
motion.

Normal Reaction R = mg cos 
= 0.5  1 cos 30o
= 4.33 N
h
R                     rou g
)
-2
i on
s
fr ict
4m                          Q(
a=                  
s in                        
mg                            c os
mg

3
horizontal
In the direction along the inclined plane, there is MOTION.
Apply F = ma
mg sin  - Q = ma
Q = mg sin - ma
Q = 0.5  1 sin 30o - .5 4
h
= 2.5 - 2.                       R                    rou g
)
.5N
-2
i on
s
fr ict
4m                         Q(
a=                  
s in                       
mg                           c os
mg

3
horizontal
At a certain inclination of the runway, the trolley can move down-
slope with uniform velocity after given a push.
Find the value  for this to occur.                  R
-2
y
m s velocit
h
Q = .5N                                                            roug
a = 0 iform                                n   )
n
fo r u        sin          Q   (f rictio
The forces along the                     mg
inclined should be
balanced                                               mg cos 
horizontal
mg sin  = Q
0.5  1 sin  = .5
 = 5.74o
At  = 5.74o , the runway is friction-compensated.

R
-2
Q = .5N                                       y
m s velocit
h
roug
mg sin 5.74o = 0.5 N        a = 0 iform                                    n   )
n
fo r u        sin            Q     (f rictio
mg

             mg cos 
horizontal

At this inclination, the friction is balanced by the down-slope
component of the weight.
In doing experiments concerning a trolley moving on a
horizontal runway, friction affects the results of the experiment.

By sloping the runway at a certain angle (usually very small)
with the horizontal, the effect of friction can be canceled. The
runway is said to be friction-compensated.

To find the inclination of the runway for friction compensation,
the runway can be slightly inclined and then a trolley on it is
given a push. If the trolley moves down the runway with
uniform velocity, it is friction-compensated. Otherwise the
inclination is adjusted again. The velocity of the trolley can be
studied using a ticker-tape timer or a motion sensor.

```
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 views: 26 posted: 3/26/2011 language: English pages: 8