NJC 02. Forces and Moments Class Exercises Solution

National Junior College Science Department | Physics 1 02 FORCES AND MOMENTS Reference notes (page 10) Class exercise The spring constant k of a spring may be determined by finding the extension of the spring and the load applied, using the apparatus shown . A student obtained the following readings. reading on the rule for the lower end of the unextended spring = 13.60 ± 0.05 cm reading on the rule for the lower end of the extended spring = 17.95 ± 0.05 cm load = 4.00 ± 0.02 N Determine k and express it with its actual uncertainty. Suggested solution Extension of spring 𝑥 = 17.95 − 13.60 = 4.35 cm Δ𝑥 = 0.05 + 0.05 = 0.1 cm 𝑘 = 𝐹 𝑥 = 4 0.0435 = 91.954 N m−1 Δ𝑘 𝑘 = Δ𝐹 𝐹 + Δ𝑥 𝑥 Δ𝑘 91.954 = 0.02 4 + 0.1 4.35 Δ𝑘 = 3 N m−1 ∴ 𝑘 = (92 ± 3) N m−1

National Junior College Science Department | Physics 2 Reference notes (page 11) Combining springs Springs can be combined in different ways, end-to-end (in series) and side-by-side (in parallel). (a) Determine the equivalent force constant of two identical springs, each of force constant k, in series and two springs in parallel, as shown below. springs in series Hint for spring in series: 1. A same force equal in magnitude to the load acts on each spring. 2. The equivalent spring will have the same amount of extension as the total extension of the two springs. springs in parallel Hint for spring in parallel: 1. The two springs must have the same amount of extension. 2. The equivalent spring will have the same amount of extension as the two springs. (b) Determine the equivalent force constant of three or more springs with difference force constants when they are combined in series and parallel. Suggested solution (a) Let F be load on the springs [Series combination] Extension of each spring = 𝐹 𝑘 𝑘′𝑥 = 𝐹 ⇒ 𝑘′ (2 × 𝐹 𝑘) = 𝐹 ⇒ 𝑘′ = 1 2 𝑘 [Parallel combination] Load on each spring = 𝐹 2 Extension of each spring = 𝐹 2𝑘 𝑘′𝑥 = 𝐹 ⇒ 𝑘′ ( 𝐹 2𝑘) = 𝐹 ⇒ 𝑘′ = 2𝑘 (b) [Series combination] Extension of each spring is 𝐹 𝑘1 , 𝐹 𝑘2 ,…, 𝐹 𝑘𝑁 Total extension of springs 𝑥 = 𝐹 𝑘1 + 𝐹 𝑘2 + ⋯ + 𝐹 𝑘𝑁 𝑘′𝑥 = 𝐹 ⇒ 𝑘′ ( 𝐹 𝑘1 + 𝐹 𝑘2 + ⋯ + 𝐹 𝑘𝑁 ) = 𝐹 ∴ 𝑘′ = ( 1 𝑘1 + 1 𝑘2 + ⋯ + 1 𝑘𝑁 ) −1 (like resistors in parallel!) [Parallel combination] Let the equivalent spring have extension x. Load 𝐹 = 𝑘1𝑥 + 𝑘2𝑥 + ⋯ 𝑘𝑁𝑥 𝑘′𝑥 = 𝐹 ⇒ 𝑘′𝑥 = 𝑘1𝑥 + 𝑘2𝑥 + ⋯ 𝑘𝑁𝑥 ∴ 𝑘′ = 𝑘1 + 𝑘2 + ⋯ 𝑘𝑁 (like resistor in series!)

National Junior College Science Department | Physics 3 Reference notes (page 18) Class exercise Draw the force diagrams for the following objects (underlined item). (a) A uniform beam hinged to a wall and supported by a wire. (b) A ladder resting on a rough wall and rough floor. (c) A uniform rod supported by a vertical string on one end and the other end resting on a smooth floor. (d) A light winch holding up a heavy box. Draw the forces in the side view diagram.