Question 38
Fig. 4.1 shows a metal
cylinder of height 4.5 cm and base area 24 cm2.
Fig. 4.1
The density of the
metal is 7900 kg m-3.
(a)
Show that the mass of the cylinder is 0.85 kg. [2]
(b)
The cylinder is placed on a plank, as shown in Fig. 4.2.
Fig. 4.2
The plank is at an
angle of 40° to the horizontal.
Calculate the pressure
on the plank due to the cylinder. [3]
(c)
The cylinder then slides down the plank with a constant acceleration
of 3.8 m s-2.
A constant frictional
force f acts on the cylinder.
Calculate the
frictional force f. [3]
Reference: Past Exam Paper – November 2015 Paper 21 Q4
Solution:
(a)
density =
mass / volume
{mass =
density × volume
Volume =
height × base area
Volume = 4.5×10-2
× 24×10-4 = 4.5 × 24 × 10-6 m3}
mass =
7900 × 4.5 × 24×10-6 = 0.85 (0.853) kg
(b)
pressure =
force / area
{The force
is the component of the weight acting perpendicularly to the surface of the
plank.
Force =
Weight × cos 40° = W cos 40°}
force = W cos 40°
{pressure
= force / area
Force = W cos
40° = mg cos 40°}
pressure =
(0.85 × 9.81 cos
40°) / (24×10-4)
pressure =
2.7 (2.66) × 103 Pa
(c)
{Resultant
force F = ma}
F = ma
{Consider
the forces acting along the surface of the plank.
Component
of weight (downward) along plank = W sin 40° = mg cos 40°
The
frictional force f opposes motion and so, acts upwards along the plank.
The
resultant force is downwards as thr cylinder slides down the plank.
Resultant
force = W sin 40° – f = ma}
W sin 40° – f = ma
0.85×9.81×sin 40° – f = 0.85 × 3.8
{5.36 – f
= 3.23}
f (= 5.36 – 3.23) = 2.1 N
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