The weightless horizontal bar in the figure below is in equilibrium. Scale B rea
ID: 1682868 • Letter: T
Question
The weightless horizontal bar in the figure below is in equilibrium. Scale B reads 4.30 kg. The distances in the figure (which is not to scale) are: D1 = 7.5 cm, D2 = 7.0 cm, and D3 = 9.0 cm. The mass of block X is 1.03 kg and the mass of block Y is 2.05 kg. Determine what the reading on scale A must be. Determine the unspecified mass of block Z. The weightless horizontal bar in the figure below is in equilibrium. Scale B reads 4.30 kg. The distances in the figure (which is not to scale) are: D1 = 7.5 cm, D2 = 7.0 cm, and D3 = 9.0 cm. The mass of block X is 1.03 kg and the mass of block Y is 2.05 kg. Determine what the reading on scale A must be. Determine the unspecified mass of block Z.Explanation / Answer
Given Data :: B = 4.3 Kg (upward) A = ? Kg (upward) x = 1.03 Kg (downward) y = 2.05 Kg (downward) z = ? Kg (downward) D1 = 7.5 cm = 0.075 m D2 = 7.0 cm = 0.07 m D3 = 9.0 cm = 0.09 m ACCORDING TO LAWS OF "PARALLEL FORCES" we know that sum of upward forces = sum of down ward forces (first law) A + B = x + y + z A + 4.3 = 1.03 + 2.05 + z z - A = 0.22 ---------------------------------------(1) & sum of clockwise moment of mass = sum of anti_clockwise moment of mass (about A) [moment of mass about A= mass * perpendicular distance from A] A*0 + B(D1+D2+D3) = xD1 + y(D1+D2) + z(D1+D2+D3) 4.3*0.235 = (1.03*0.075) + (2.05*0.145) + z*0.235 1.0105 = 0.07725 + 0.29725 + z*0.235 z*0.235 = 0.636 z = 2.71 Kg ----------------------(2) substitute equation (2) in (1) to get the value of A 2.71 - A = 0.22 A = 2.71 - 0.22 = 2.49 Kg So A = 2.49 Kg & z = 2.71 KgRelated Questions
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