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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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question.dat
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1998-02-10
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[question1]
type:1
image:9g19
caption:\
Which one of the four forces below will exactly balance the force on \
this mass?<p>
correct:c
wrong1:b
wrong2:a
wrong3:d
feedback:\
To balance the first force, the second force must be equal and \
opposite.<p>
[question2]
type:2
caption:\
Your weight (the force of gravity on your body) acts vertically \
downwards. Which one of the following forces balances your weight so \
that you do not fall through the floor?<p>
correct:The contact (reaction) force on your feet from the floor.
wrong1:The force of gravity on the Earth from your body.
wrong2:The contact (reaction) force on the floor from your feet.
wrong3:The force of gravity on your body from the Moon.
feedback:\
When you are standing stationary on the floor, your weight is balanced \
by an equal and opposite contact force on your feet from the floor.<p>
[question3]
type:2
caption:\
When all the forces on an object are in equilibrium, they add together \
to produce a resultant force which is:<p>
correct:Zero.
wrong1:Less than the object's weight.
wrong2:Equal to the object's weight.
wrong3:Greater than the object's weight.
feedback:\
To produce equilibrium, all the forces on an object must balance. They \
add together to produce a zero resultant force.<p>
[question4]
type:1
image:9g20
caption:\
Each of the three monkeys in the diagram weighs 100 N. What is the \
tension force in the rope?<p>
correct:300 N
wrong1:Zero
wrong2:100 N
wrong3:200 N
feedback:\
The total weight of the monkeys on the rope is 300 N. If the monkeys \
are stationary, this force must be balanced by an equal and opposite \
force from the rope. The tension in the rope is therefore 300 N.<p>
[question5]
type:1
image:9g21
caption:\
This man is holding a 300 N load stationary. He weighs 800 N. What is \
the friction force on the soles of his shoes from the floor?<p>
correct:300 N
wrong1:Zero
wrong2:500 N
wrong3:1 100 N
feedback:\
The man is being pulled horizontally by the 300 N load. If he is to \
remain stationary, the pulling force must be balanced by an equal and \
opposite force. This force is provided by a 300 N friction force from \
the floor which acts on the soles of his feet.<p>
[question6]
type:1
image:9g22
caption:\
A box is resting on a sloping plank. Two of the forces acting on the \
box are shown - its weight acting vertically downwards and the contact \
force acting at right angles to the plank's surface. These forces are \
balanced by a third force. Which of the forces below shows the \
direction of this third force?<p>
correct:d
wrong1:a
wrong2:b
wrong3:c
feedback:\
The box is kept in equilibrium by the friction force acting up the \
plank.<p>
[question7]
type:3
caption:\
Which one of the masses below is not being kept in equilibrium by the \
forces acting upon it?<p>
correct:9g23b
wrong1:9g23a
wrong2:9g23c
wrong3:9g23d
feedback:\
The two forces acting upon mass B are equal, but they are not \
opposite, so they do not balance.<p>
[question8]
type:2
caption:\
A 100 kg weightlifter is standing stationary holding a 50 kg dumb-bell \
above his head. Which of the following forces is the greatest?<p>
correct:The contact force of the floor on the weightlifter's feet.
wrong1:The force of gravity on the weightlifter's body.
wrong2:The force of gravity on the dumb-bell.
wrong3:The contact force of the dumb-bell on the weightlifter's hands.
feedback:\
The upward contact force on the weightlifter's feet from the floor \
must balance both the weightlifter's weight and the weight of the \
dumb-bell, so it is the greatest of the four forces.<p>
[question9]
type:3
caption:\
Four different mechanics are using spanners to tighten nuts. Which \
mechanic is using the greatest turning force to tighten the nut?<p>
correct:9g24c
wrong1:9g24a
wrong2:9g24b
wrong3:9g24d
feedback:\
The moment or turning effect of a force is the force multiplied by its \
perpendicular distance from the pivot. The four moments are:<p>\
a) 400 x 0.2 = 80 Nm<p>\
b) 300 x 0.3 = 90 Nm<p>\
c) 250 x 0.4 = 100 Nm<p>\
d) 150 x 0.5 = 75 Nm<p>\
Therefore, mechanic C is tightening the nut with the greatest turning \
force.<p>
[question10]
type:1
image:9g25
caption:What force must you apply to the handle to raise the bucket?<p>
correct:40 N
wrong1:20 N
wrong2:100 N
wrong3:400 N
feedback:\
Apply the principle of moments to find the force needed to just raise \
the bucket. Let the force be <I>F</I>.<p>\
anticlockwise moment = clockwise moment<p>\
100 x 0.16 = <I>F</I> x 0.40<p>\
<img src="sa9q10a" align=center><p>\
<img src="sa9q10b" align=center><p>\
<center>= 40 N.</center><p>
[question11]
type:1
image:9g26
caption:\
What is the size of the force that must act at X to balance the \
beam?<p>
correct:4.0 N
wrong1:20 N
wrong2:30 N
wrong3:40 N
feedback:\
Apply the principle of moments to find the force needed to balance the \
beam. Let the force be <I>F</I>.<p>\
anticlockwise moment = clockwise moment<p>\
10 x 0.40 = <I>F</I> x 1.0<p>\
<img src="sa9q11a" align=center><p>\
<center>= 4.0 N.</center><p>
[question12]
type:1
image:9g27
caption:\
What is the size of the force that must act at X to balance the \
beam?<p>
correct:15 N
wrong1:10 N
wrong2:40 N
wrong3:20 N
feedback:\
Apply the principle of moments to find the force needed to balance the \
beam. Let the force be <I>F</I>.<p>\
anticlockwise moment = clockwise moment<p>\
20 x 1.0 = 10 x 0.50 + <I>F</I> x 1.0<p>\
20 = 5.0 + <I>F<p>\
</I><p>\
F = 20 - 5.0<p>\
<center>= 15 N.</center><p>
[question13]
type:1
image:9g28
caption:\
What is the size of the force that must act at X to balance the \
beam?<p>
correct:6.0 N
wrong1:50 N
wrong2:30 N
wrong3:10 N
feedback:\
Apply the principle of moments to find the force needed to balance the \
beam. Let the force be <I>F</I>.<p>\
anticlockwise moment = clockwise moment<p>\
20 x 0.50 + 15 x 0.40 = 10 x 1.0 + <I>F</I> x 1.0<p>\
10 + 6.0 = 10 + <I>F<p>\
</I><p>\
<I>F</I> = 16 - 10 N<p>\
<center>= 6.0 N.</center><p>
[question14]
type:1
image:9g29
caption:At which point must the force X be placed to balance the beam?<p>
correct:B
wrong1:A
wrong2:C
wrong3:D
feedback:\
Apply the principle of moments. Let the distance of the force from the \
pivot be <I>d</I>. Without the extra force, the beam will rotate \
clockwise. To balance the beam the upward force must act to the right \
of the pivot, in order to produce an anticlockwise moment.<p>\
anticlockwise moment = clockwise moment<p>\
100 x <I>d</I> = 50 x 0.50<p>\
<img src="sa9q14a" align=center><p>\
<center>= 0.25 m</center><p>\
The force must therefore be placed at B.<p>
[question15]
type:1
image:9g30
caption:At which point must the force X be placed to balance the beam?<p>
correct:C
wrong1:A
wrong2:B
wrong3:D
feedback:\
Apply the principle of moments. Let the distance of the force from the \
pivot be <I>d</I>. Without the extra force, the beam will rotate \
clockwise. To balance the beam the upward force must act to the right \
of the pivot, in order to produce an anticlockwise moment.<p>\
anticlockwise moment = clockwise moment<p>\
10 x 0.50 + 20 x <I>d</I> = 20 x 0.50<p>\
5.0 + 20<I>d </I>= 10<p>\
20<I>d</I> = 10 - 5.0<p>\
<img src="sa9q15a" align=center><p>\
<center>= 0.25 m</center><p>\
The force must therefore be placed at C.<p>
[question16]
type:1
image:9g31
caption:\
A large mass and a smaller mass are fixed to either end of a light \
metal rod. At which point is the centre of gravity of the whole object \
located?<p>
correct:B
wrong1:A
wrong2:C
wrong3:D
feedback:\
The centre of gravity is the point at which the object will balance. \
This point lies between the two masses, and will be closer to the \
larger mass than the smaller one (you can confirm this by applying the \
principle of moments about the centre of gravity). The object's centre \
of gravity is therefore at point B.<p>
[question17]
type:1
image:9g32
caption:\
The weight of the uniform beam in the diagram is 200 N. What is the \
tension in the string?<p>
correct:100 N
wrong1:50 N
wrong2:200 N
wrong3:400 N
feedback:\
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>\
For equilibrium:<p>\
sum of anticlockwise moments = sum of clockwise moments<p>\
<i>T</i> x 2.0 = 200 x 1.0<p>\
Therefore,<p>\
<img src="sa9q17a" align=center><p>\
<center>= 100 N.</center><p>
[question18]
type:1
image:9g33
caption:\
The diagram shows a wheel on an axle in four different positions. In \
which position is this object in unstable equilibrium?<p>
correct:b
wrong1:a
wrong2:c
wrong3:d
feedback:\
In unstable equilibrium, the centre of gravity falls when the object \
is displaced so the object topples. This is the situation in position \
b.<p>
[question19]
type:1
image:9g33
caption:\
The diagram shows a wheel on an axle in four different positions. In \
which position is this object in neutral equilibrium?<p>
correct:a
wrong1:b
wrong2:c
wrong3:d
feedback:\
In neutral equilibrium, the centre of gravity stays at the same height \
when the object is displaced so the object rolls. This is the \
situation in position a.<p>
[question20]
type:1
image:9g33
caption:\
The diagram shows a wheel on an axle in four different positions. In \
which position is this object not in equilibrium?<p>
correct:d
wrong1:a
wrong2:b
wrong3:c
feedback:\
When an object is not in equilibrium, the forces on it do not balance \
and its motion changes. This is the situation in position d, where the \
unbalanced weight of the axle will make the wheel fall over.<p>