Q1.
Figure 1 shows a garden gate with a pulley system designed to close the gate.
Figure 1
The pulley system raises weight A when the gate is opened. When the gate is released, A
falls. The horizontal cable C passes over pulley R. The tension in cable C causes the gate
to close.
Weight A is a solid cylinder with the following properties:
diameter = 4.8 × 10–2 m
length = 0.23 m
weight = 35 N
The table below gives the density of three available materials.
Material Density / kg m–3
concrete 2.4 × 103
iron 7.8 × 103
brass 8.6 × 103
(a) Deduce which one of the three materials is used for A.
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(3)
Figure 2 shows the pulley arrangement when the gate is closed.
Figure 2
Pulleys P and M are frictionless so that the tension in the rope attached to A is equal to
the weight of A.
A weighs 35 N and the weight of moveable pulley M is negligible.
(b) Calculate the tension in the horizontal cable C when the gate is closed.
tension = _______________ N
(2)
(c) Pulley M is pulled to the left as the gate is opened.
Explain why this increases the tension in the horizontal cable C.
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(2)
(d) Figure 3 shows a plan view with the gate open. The horizontal cable C passes over
pulley R and is attached to the door at D.
The angle between the door and the horizontal cable C is 12°.
The horizontal distance between the hinge and D is 0.95 m.
Figure 3
The tension in the horizontal cable C is now 41 N.
Calculate the moment of the tension about the hinge.
moment = _______________ N m
(2)
(e) The same system is attached to an identical gate with stiffer hinges. Now the system
does not supply a sufficiently large moment to close the gate.
Discuss two independent changes to the design to increase the moment about the
hinges due to horizontal cable C.
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2 _________________________________________________________________
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(4)
(Total 13 marks)
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Mark schemes
Q1.
(a) Volume of A = area × length = 4.16 × 10−4 m3
OR
Mass of A = W/g = 3.6 kg ✔
Condone POT error in MP1
Use density equation ✔
Do not allow use of weight in density equation
Compares
a calculated property of brass (e.g. weight, length or diameter) with A
OR
the calculated density of A with density of brass
Do not accept 8.3 × 103 for density of A .
OR
the calculated mass of A with the calculated mass of brass
Award zero marks for an unsupported answer “Brass”
and therefore brass ✔
Only award MP3 if answer “brass” given.
Example:
Volume of A = area × length = 4.16 × 10−4 m3 ✔
Mass if brass = density × volume = 3.58 kg ✔
Weight = 3.58 × 9.81 = 35 N (which is weight of A) and therefore brass
is correct.
3
(b) Use of T = (35) cos 55 ✔
2 × their T (= 40 N ) ✔
2
(c) Angle (to horizontal) decreases ✔
(Weight/tension in rope remains constant at 35 N )
So horizontal components (from tension in rope) increase ✔
Do not award MP2 if answer suggests that tension in rope
increases.
(Therefore tension in cable must increase)
Do not allow “tension increases” for credit.
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2
(d) Component of the force at right angle to door
= 41 cos (90−12) / 41 sin (12)
= 8.5 N ✔
Moment = 8.5 × 0.95 = 8.1 (N m) ✔
Alternative:
Perpendicular distance = 0.95 sin (12)
= 0.198 m ✔
Moment = 41 × 0.198 = 8.1 ✔
Allow ecf from their value of weight component .
Allow ecf from their value of perpendicular distance.
(Calculator value is 8.098 160 3)
Award zero marks for simply multiplying 41 N × 0.95 m.
2
(e) ALTERNATIVE 1
Increase weight / density / mass / volume of A ✔
Increases tension (and therefore moment) ✔
ALTERNATIVE 2
Position pulley R further (out) from gate hinges / increase diameter of pulley R. ✔
Increases angle and therefore bigger perpendicular component (and
therefore moment). ✔
Any 2 alternatives
ALTERNATIVE 3
Decrease angle of rope eg by putting P and fixed point closer together /
further to right. ✔
Increases tension (and therefore moment). ✔
If more than two answers given, mark first two. Ignore the 1
and 2 in answer lines.
ALTERNATIVE 4
Move D further from hinge/R OR make C longer. ✔
Increases perpendicular distance (and therefore moment). ✔
4
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