Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm2.

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