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Investigating Forces & Motion
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Investigating Forces and Motion (1998)(Granada Learning).iso
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topic9
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example.dat
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1998-02-10
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59 lines
[general]
[page1]
type:1
caption:\
These diagrams show four objects being acted on by forces. None of the \
objects are in equilibrium. Which of the forces beneath will balance \
the existing forces? Drag the arrows to match the extra forces to the \
objects.<p>
feedback:\
Correct. In each case the extra force must exactly balance the effect \
of the forces that are already acting. Any turning effect of the \
forces must be balanced, as well as their tendency to produce \
acceleration in a straight line.<p>
source:9ex1b1, 9ex1c1, 9ex1a1, 9ex1d1
target:9ex1a, 9ex1b, 9ex1c, 9ex1d
[page2]
type:0
caption:\
A uniform 2.0 m beam with a weight of 50 N is pivoted at one end. It \
is held horizontal by a vertical string fixed to the other end. What \
is the tension in the string?<p>\
The beam is uniform. This means that its centre of gravity is at its \
mid point, 1.0 m from the pivot. The beam is in equilibrium so the \
principle of moments can be applied.<p>\
<img src="9ex2" align=center><p>\
For equilibrium<p>\
sum of anticlockwise moments = sum of clockwise moments<p>\
Taking moments about the pivot,<p>\
<i>T</i> x 2.0 = 50 x 1.0<p>\
Therefore, <img src="r9ex2" align=center><p>\
<center>= 25 N</center><p>
[page3]
type:0
caption:\
A uniform beam is pivoted at the centre. A 20 N weight is placed 0.4 m \
from the pivot. Where must a 5.0 N weight be placed to balance the \
beam?<p>\
The beam is uniform and pivoted at its centre. This means that the \
turning moment of its own weight about the pivot is zero.<p>\
To achieve balance the second weight must be placed on the opposite \
side of the pivot to the first weight. Let its distance from the pivot \
be <I>X</I>.<p>\
When the beam is balanced the principle of moments can be applied<p>\
For equilibrium<p>\
sum of anticlockwise moments = sum of clockwise moments<p>\
Taking moments about the pivot,<p>\
20 x 0.4 = 5.0 x <I>X<p>\
</I><p>\
Therefore,<p>\
<img src="r9ex3" align=center><p>\
<center>= <U>1.6 m</center><p>\
</U><p>\
The second weight must be placed 1.6 m from the pivot on the opposite \
side of the beam.<p>