Problems on lamis theorem.pdf
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This document contains solutions to problems involving Lami's theorem. Lami's theorem relates tensions in strings or ropes that meet at a single point. The document contains solutions for determining tensions in ropes supporting weights, tensions between blocks suspended from ropes, and pressures between cylinders resting on surfaces. Solutions involve drawing free body diagrams and applying Lami's theorem to relate angles and tensions or pressures at contact points. The document provides step-by-step workings to model systems of connected objects in static equilibrium and solve for unknown tensions, pressures and weights.
Problems on lamis theorem.pdf
- 1. Prepared by: Prof. V.V. Nalawade 1 Problems on Lami’s Theorem Q1. Find the tension in each rope in fig. ANS: Given data: W = 100 kg = 100*9.81= 981 N θ = tan−1 4 3 = 53.13ᵒ Step 1: Consider the FBD of point C General Calculation : Angle between W and TBC = 90 – 30 = 60ᵒ Angle between TBC & TAC = 30 + 90 + (90 - 53.13) =156.87ᵒ Angle between TAC & W = 53.13 + 90 =143.13ᵒ Step 2: Apply Lami’s Theorem , 981 sin 156.87° = 𝑇𝐴𝐶 sin 60° = 𝑇𝐵𝐶 sin 143.13° Therefore, TAC = 2162.76 N TBC = 1438.41 N
- 2. Prepared by: Prof. V.V. Nalawade 2 Q2. Block P = 5kg and block Q of mass m kg is suspended through the chord is in the equilibrium position as shown in fig. Determine the mass of block Q. ANS: Given data: i. P = 5 kg = 5*9.81 = 49.05 N ii. Q = m kg iii. θ = tan−1 4 3 = 53.13ᵒ CASE 1: CONSIDER JOINT B Step 1: Consider the FBD of Joint B FBD of Joint B General Calculation : Angle between P and TBC = 90 + 30 = 120ᵒ Angle between TBC & TAB= 180 – 53.13 - 30 =96.87ᵒ Angle between TAB & P = 53.13 + 90 =143.13ᵒ Step 2: Apply Lami’s Theorem , 49.05 sin 96.87° = 𝑇𝐴𝐵 sin 120° = 𝑇𝐵𝐶 sin 143.13° Therefore, TAB = 42.79 N
- 3. Prepared by: Prof. V.V. Nalawade 3 TBC = 29.64 N CASE 2: CONSIDER JOINT C Step 1: Consider the FBD of Joint C FBD of Joint C General Calculation : Angle between Q and TCD = 90 + 70 = 160ᵒ Angle between TCD & TBC= 180 – 70 + 30 =140ᵒ Angle between TBC & Q = 90-30 =60ᵒ Step 2: Apply Lami’s Theorem , 𝑄 sin 140° = 29.64 sin 160° = 𝑇𝐶𝐷 sin 60° Therefore, Q = 55.70 N = 𝟓𝟓.𝟕𝟎 𝟗.𝟖𝟏 = 𝟓. 𝟔𝟖 𝒌𝒈 TCD = 75.05 N Q3. A light string ABCDE whose extremity A is fixed, has weights W1 and W2 attached to it at B and C. It passes round peg at D carrying a weight of 300 N at the free end E as shown in fig. if in the equilibrium position, BC is horizontal and AB and CD make 150ᵒ and 120ᵒ with BC, find (i) Tension in the portion AB, BC and CD of the string and (ii) magnitude of W1 and W2.
- 4. Prepared by: Prof. V.V. Nalawade 4 ANS: Given data: i. Weight at E = 300 N ii. TCD = TDE = 300 N CASE 1: CONSIDER JOINT B Step 1: Consider the FBD of Joint B & c and apply Lami’s theorem at C, Fig: FBD of Joint C 𝟑𝟎𝟎 𝑺𝒊𝒏 𝟗𝟎 = 𝑻𝑩𝑪 𝑺𝒊𝒏 𝟏𝟓𝟎 = 𝑾𝟐 𝑺𝒊𝒏 𝟏𝟐𝟎 Hence, TBC = 𝟑𝟎𝟎 𝑺𝒊𝒏 𝟗𝟎 𝒙 𝑺𝒊𝒏 𝟏𝟓𝟎 = 150N Similarly W2 = 259.81 N CASE 1: CONSIDER JOINT B Step 2: Consider the FBD of Joint B and apply Lami’s theorem at B,
- 5. Prepared by: Prof. V.V. Nalawade 5 Fig: FBD of Joint B 𝑻𝑨𝑩 𝑺𝒊𝒏 𝟗𝟎 = 𝑻𝑩𝑪 𝑺𝒊𝒏 𝟏𝟐𝟎 = 𝑾𝟏 𝑺𝒊𝒏 𝟏𝟓𝟎 We know that, TBC = 150N Therefore, TAB = 𝟏𝟓𝟎 𝑺𝒊𝒏 𝟏𝟐𝟎 𝒙 𝑺𝒊𝒏 𝟗𝟎 = 173.21 N Similarly, W1 = 86.60 N Therefore, W1 = 86.60 N and W2 = 259.81 N respectively. EXERCISE : 1
- 6. Prepared by: Prof. V.V. Nalawade 6 Q.4 Two Cylinders P & Q rest in a channel as shown in fig. The cylinder P has diameter of 100 mm and weighs 200 N, whereas the cylinder Q has diameter of 180 mm and weighs 500 N. SOL: Given data: i. Diameter of Cylinder P = 100 mm ii. Weight of Cylinder P = 200 N iii. Diameter of Cylinder Q = 180 mm iv. Weight of Cylinder Q = 500 N v. Width of channel = 180 mm. Step 1: Consider the FBD of Cylinder P Fig: FBD of cylinder P General Calculation : From geometry of the figure, we can find that ED = radius of cylinder P = 100 2 = 50 𝑚𝑚 BF = radius of cylinder Q = 180 2 = 90 𝑚𝑚
- 7. Prepared by: Prof. V.V. Nalawade 7 And ∆ BCF = 60 ᵒ Hence, CF = 𝐵𝐹 𝑇𝑎𝑛 60 = 52 𝑚𝑚 FE = BG = 180 – 52 – 50 = 78 mm And AB = 50 + 90 = 140 mm Hence ∆ ABG = COS-1 𝐵𝐺 𝐴𝐵 = 78 100 = 56.14° Step 2: Apply Lami’s theorem at A, 𝑹𝟏 𝑺𝒊𝒏 ( 𝟗𝟎 + 𝟓𝟔. 𝟏𝟒) = 𝑹𝟐 𝑺𝒊𝒏 𝟗𝟎 = 𝟐𝟎𝟎 𝑺𝒊𝒏 (𝟏𝟖𝟎 − 𝟓𝟔. 𝟏𝟒) Hence, R1 =134.19 N & R2 = 240.85 N Step 3: Consider the FBD of Cylinder Q Fig: FBD of cylinder Q Step 4: Apply Lami’s theorem at B, 𝑹𝟑 𝑺𝒊𝒏 (𝟗𝟎+𝟓𝟔.𝟏𝟒) = 𝟐𝟒𝟎.𝟖 𝑺𝒊𝒏 (𝟔𝟎) = 𝑹𝟒−𝟓𝟎𝟎 𝑺𝒊𝒏(𝟏𝟖𝟎+𝟑𝟎−𝟓𝟔.𝟏𝟒) Hence, R3 = 155 N & R4 = 622.50 N Q5. Three cylinders weighing 100 N each and of 80 mm diameter are
- 8. Prepared by: Prof. V.V. Nalawade 8 placed in a channel of 180 mm width as shown in fig. Determine the pressure exerted by i) the cylinder A on B at the point of contact ii) the cylinder B on the base & iii) the cylinder B on the wall. SOL: Given data: i. weight of each Cylinder = 100 N ii. Diameter of Cylinder = 80 mm and iii. Width of channel = 180 mm Case 1: Pressure exerted by the cylinder A on the cylinder B Let, R1 = pressure exerted by the cylinder A on B. Step 1: Consider the FBD of cylinder A Fig: FBD of cylinder A General Calculation : From geometry of the triangle OPS, we can find that OP = 40 + 40 = 80 mm
- 9. Prepared by: Prof. V.V. Nalawade 9 PS = 90 – 40 = 50 mm Therefore, ∆ POS = Sin-1 𝑃𝑆 𝑂𝑃 = 50 80 = 38.68° Step 2: Apply Lami’s theorem at O, 𝑹𝟏 𝑺𝒊𝒏 ( 𝟏𝟒𝟏. 𝟑) = 𝑹𝟐 𝑺𝒊𝒏 𝟏𝟒𝟏. 𝟑 = 𝟏𝟎𝟎 𝑺𝒊𝒏 (𝟕𝟕. 𝟒) Hence, R1 = R2 = 64 N Case 2: Pressure exerted by the cylinder C on the base Step 3: Consider the FBD of cylinder B Fig: FBD of cylinder A General Calculation : Let, R3 = Pressure exerted by the cylinder B on the wall, And R4 = Pressure exerted by the cylinder B on the base. Step 4: Apply Lami’s theorem at P, 𝟔𝟒 𝑺𝒊𝒏 ( 𝟗𝟎) = 𝑹𝟑 𝑺𝒊𝒏 𝟏𝟖𝟎 − 𝟑𝟖. 𝟕 = (𝑹𝟒 − 𝟏𝟎𝟎) 𝑺𝒊𝒏 (𝟗𝟎 + 𝟑𝟖. 𝟕) Hence, R3 = 40 N & R4 = 150 N EXERCISE : 2 Q1. A right circular cylinder of diameter 40 cm, open at both
- 10. Prepared by: Prof. V.V. Nalawade 10 ends, rests on a smooth horizontal plane. Inside the cylinder, there are two sphere having weights and radii as given. Find the minimum weight of the cylinder for which it will not tip over. Ans : R1 = R2 = 212.16 N & W = 300 N Q2. Three cylinders are piled up in a rectangular channel as shown in fig. determine the reactions at point 6 between the cylinder A and the vertical wall of the channel. Ans : R6 = 784.8 N Q3. Two identical rollers, each of weight 50 N, are supported by an inclined and a vertical wall as shown in fig find the reactions at the points of supports A,B & C. Assume all the surfaces to be smooth.
- 11. Prepared by: Prof. V.V. Nalawade 11 Ans : Ra = 43.3 N , Rb = 72 N & Rc = 57.5 N