Question 19A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm.(i) Find the value of m.(ii) To which side the rule will tilt if the mass m is moved to the mark 10 cm?(iii) What is the resultant moment now?(iv) How can it be balanced by another mass 50 g?

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Accepted Answer

(i) As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment100g×(50−40)cm=m×(40−20)cm⇒100×10=m×20⇒m=50g100g \times (50-40)cm = m \times (40-20)cm \[0.5em] \Rightarrow 100 \times 10 = m \times 20 \[0.5em] \Rightarrow m = 50g \[0.5em]100g×(50−40)cm=m×(40−20)cm⇒100×10=m×20⇒m=50g(ii) When the mass m is shifted to mark 10cm , it results in rule being shifted on the side of mass m in anticlockwise direction.(iii) Anticlockwise moment is produced when a mass of m grams is moved towards the mark of 10cm.100gf×(50−40)cm=1000gfcm100gf \times (50-40)cm = 1000gfcm100gf×(50−40)cm=1000gfcmTherefore, the resultant moment will be1500gfcm−1000gfcm=500gfcm(anticlockwise)1500gfcm - 1000 gfcm = 500gfcm (anticlockwise)1500gfcm−1000gfcm=500gfcm(anticlockwise)(iv) As we know, the principle of moments states thatAnticlockwise moment = Clockwise momentSo,100gf×(50−40)cm+50gf×d=50gf×(40−10)cm⇒1000gfcm+50gf×d=1500gfcm⇒50gfd=500gfcm⇒d=10cm100gf \times (50-40)cm + 50gf \times d = 50gf \times (40-10)cm \[0.5em] \Rightarrow 1000gfcm + 50gf \times d = 1500gfcm \[0.5em] \Rightarrow 50gfd = 500gfcm \[0.5em] \Rightarrow d = 10 cm100gf×(50−40)cm+50gf×d=50gf×(40−10)cm⇒1000gfcm+50gf×d=1500gfcm⇒50gfd=500gfcm⇒d=10cmHence, we can balance 50gm at 50cm

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