Q#38 (Past Exam Paper – November 2015 Paper 21 Q4)
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]
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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