HCI 04 Forces Solutions (Lecture Examples)

Example 1 A spring, obeying Hooke’s Law, has an unstretched length of 50 mm and a spring constant of 400 N m-1. What is the tension in the spring when its overall length is 70 mm? The extension of the spring is x = 70 – 50 = 20 mm = 0.020 m The tension in the spring is T = k x = (400) (0.020) = 8.0 N Convert to base SI units

a) What is the pressure of the point X in mm Hg? b) What is the pressure of the gas in mm Hg? c) If the manometer is disconnected from the gas container, what will be the vertical heights of the mercury levels from the point X in the right and left limbs of the manometer? Example 2 The diagram on the right shows a mercury manometer connected to a gas container. The atmospheric pressure is 760 mm Hg and the point X is at the bottom of the U-tube. a) (pressure at point X) = (pressure due to fluid from B to X) + (atmospheric pressure) = (34 mm Hg) + (760 mm Hg) = 794 mm Hg Pressure in mm Hg is defined as the pressure due to a column of so many mm of mercury

a) (pressure at point X) = 794 mm Hg b) (pressure of the gas) + (pressure due to fluid from A to X) = (pressure at point X) (pressure of the gas) = (pressure at point X) - (pressure due to fluid from A to X) = (794 mm Hg) – (18 mm Hg) = 776 mm Hg a) What is the pressure of the point X in mm Hg? b) What is the pressure of the gas in mm Hg? Example 2 The diagram on the right shows a mercury manometer connected to a gas container. The atmospheric pressure is 760 mm Hg and the point X is at the bottom of the U-tube.

c) If the manometer is disconnected from the gas container, what will be the vertical heights of the mercury levels from the point X in the left and right limbs of the manometer? Example 2 The diagram on the right shows a mercury manometer connected to a gas container. The atmospheric pressure is 760 mm Hg and the point X is at the bottom of the U-tube. c) If the manometer is disconnected from the gas container, both open ends of the tube will be at atmospheric pressure: the pressure at the top of each column will be the same. Thus, the mercury will distribute so that it reaches the same height in both limbs. The height will be (18 + 34) / 2 = 26 mm.

The upthrust acting on the floating iceberg is equal to its weight. From Archimedes’ Principle, upthrust = weight of water displaced by the submerged part of the iceberg U = W mw g = mi g mw = mi ρw Vw = ρi Vi Vw / Vi = ρi / ρw = (0.917) / (1.025) = 0.895 Since the volume of the displaced water is 89.5% of the volume of the iceberg, 89.5% of the iceberg is submerged in the sea water. Example 3 Icebergs are often a danger to ships, since most of an iceberg‘s volume lies beneath the surface of the water. One can only see the “tip of the iceberg.” Find the percentage of a floating iceberg that is submerged. (density of ice ρi = 0.917 g cm-3, density of sea water ρw = 1.025 g cm-3)

Four forces act on a particle such that it is in equilibrium. Three of the forces are shown in the diagram. Find the magnitude and direction of the unknown force F that maintains the equilibrium. Example 4 Thus, the horizontal component of the resultant of the three forces RX is given by, (taking rightwards positive) RX = (-7.0 cos 60°) + (8.5 cos 45°) = 2.51 N The vertical component of the resultant of the three forces RY is given by, (taking upwards as positive) RY = 11.0 - (7.0 sin 60°) - (8.5 sin 45°) = -1.07 N

For the particle to be in equilibrium, the force F must be equal in magnitude and opposite in direction to the vector sum of the three forces. Fhorizontal = - RX = -2.51 N (i.e. 2.51 N to the left) Fvertical = - RY = 1.07 N (i.e. 1.07 N upwards) Using Pythagoras’ theorem, the magnitude of F is 2.73 N. tan θ = 1.07 / 2.51, so θ = 23° Hence, F has a magnitude of 2.73 N and it is pointing at 67° anticlockwise from the force of 11.0 N. Four forces act on a particle such that it is in equilibrium. Three of the forces are shown in the diagram. Find the magnitude and direction of the unknown force F that maintains the equilibrium. Example 4 θ2.51 N 1.07 NF

In the diagram on the right, a body S of weight W hangs vertically by a thread tied at Q to the string PQR. If the system is in equilibrium, what is the tension in section PQ?. Example 5 For the system to be in equilibrium, the vertical component of the tension in section QR must be equal to the tension in the section QS, which in turn must be equal to the weight W. TQR,vertical = W TQR sin 30° = W TQR = W / (sin 30°) = 2.0W For the system to be in equilibrium, the tension in section PQ must be equal to the horizontal component of the tension in section QR. TPQ = TQR (cos 30°) = 1.73W

A uniform rod of weight F and length L is pivoted on one end O. A force 2F is applied at a quarter of its length from O. Find the magnitude of the moment about point O. Example 6 The moment about O due to the rod’s weight is (F) (L/2) = 0.500FL (in the clockwise direction) The moment about O due to the force 2F is (2F) (L/4) (sin 35°) = 0.287FL (in the anticlockwise direction) Hence, the net moment about O is (0.500FL) – (0.287FL) = 0.213FL (in the clockwise direction)

By the principle of moments, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about that same point. Taking moments about O, the clockwise moment due to the weight τclockwise = (25 N) (L/4) Hence, the anticlockwise moment about O due to F is τanticlockwise = (F) (3L/4) = (25 N) (L/4) Simplifying, we find F = (25 N) / 3 = 8.33 N A uniform rod of weight 25 N and length L is balanced on a knife edge at O. Find the unknown force F required to maintain rotational equilibrium. Example 7 Moments about O must be equal in magnitude