ক) AI-তৈরি নমুনা উত্তর
ধরি, প্রথম বল দুটি P 1 = 3 P P_1 = 3P P 1 = 3 P P 2 = 4 P P_2 = 4P P 2 = 4 P θ \theta θ P 3 = 5 P P_3 = 5P P 3 = 5 P
যেহেতু বল তিনটি সাম্যাবস্থায় আছে, তাই যেকোনো দুটির লব্ধি তৃতীয় বলটির সমান ও বিপরীতমুখী হবে।3 P 3P 3 P 4 P 4P 4 P R = 5 P R = 5P R = 5 P
আমরা জানি,R 2 = P 1 2 + P 2 2 + 2 P 1 P 2 cos θ R^2 = P_1^2 + P_2^2 + 2P_1P_2\cos\theta R 2 = P 1 2 + P 2 2 + 2 P 1 P 2 cos θ ( 5 P ) 2 = ( 3 P ) 2 + ( 4 P ) 2 + 2 ( 3 P ) ( 4 P ) cos θ (5P)^2 = (3P)^2 + (4P)^2 + 2(3P)(4P)\cos\theta ( 5 P ) 2 = ( 3 P ) 2 + ( 4 P ) 2 + 2 ( 3 P ) ( 4 P ) cos θ 25 P 2 = 9 P 2 + 16 P 2 + 24 P 2 cos θ 25P^2 = 9P^2 + 16P^2 + 24P^2\cos\theta 25 P 2 = 9 P 2 + 16 P 2 + 24 P 2 cos θ 25 P 2 = 25 P 2 + 24 P 2 cos θ 25P^2 = 25P^2 + 24P^2\cos\theta 25 P 2 = 25 P 2 + 24 P 2 cos θ 24 P 2 cos θ = 0 24P^2\cos\theta = 0 24 P 2 cos θ = 0 P ≠ 0 P \neq 0 P = 0 cos θ = 0 = cos 90 ∘ \cos\theta = 0 = \cos 90^\circ cos θ = 0 = cos 9 0 ∘ ∴ θ = 90 ∘ \therefore \theta = 90^\circ ∴ θ = 9 0 ∘
অতএব, প্রথম বল দুটি পরস্পর লম্ব। (প্রমাণিত)
খ) AI-তৈরি নমুনা উত্তর
চিত্রানুযায়ী, সংযোগ বিন্দুতে ক্রিয়াশীল বলত্রয় হলো T 1 T_1 T 1 T 2 T_2 T 2 W W W T 1 T_1 T 1 30 ∘ 30^\circ 3 0 ∘ T 2 T_2 T 2 α \alpha α
সুতরাং:
T 1 T_1 T 1 W W W = 180 ∘ − 30 ∘ = 150 ∘ = 180^\circ - 30^\circ = 150^\circ = 18 0 ∘ − 3 0 ∘ = 15 0 ∘ T 2 T_2 T 2 W W W = 180 ∘ − α = 180^\circ - \alpha = 18 0 ∘ − α T 1 T_1 T 1 T 2 T_2 T 2 = 30 ∘ + α = 30^\circ + \alpha = 3 0 ∘ + α
লামির সূত্রানুসারে:T 2 sin ( 180 ∘ − 30 ∘ ) = W sin ( 30 ∘ + α ) \frac{T_2}{\sin(180^\circ - 30^\circ)} = \frac{W}{\sin(30^\circ + \alpha)} s i n ( 18 0 ∘ − 3 0 ∘ ) T 2 = s i n ( 3 0 ∘ + α ) W T 2 sin 30 ∘ = W sin ( 30 ∘ + α ) \frac{T_2}{\sin 30^\circ} = \frac{W}{\sin(30^\circ + \alpha)} s i n 3 0 ∘ T 2 = s i n ( 3 0 ∘ + α ) W T 2 = W sin 30 ∘ sin ( 30 ∘ + α ) T_2 = \frac{W \sin 30^\circ}{\sin(30^\circ + \alpha)} T 2 = s i n ( 3 0 ∘ + α ) W s i n 3 0 ∘
T 2 T_2 T 2 sin ( 30 ∘ + α ) \sin(30^\circ + \alpha) sin ( 3 0 ∘ + α ) 1 1 1 ∴ sin ( 30 ∘ + α ) = 1 = sin 90 ∘ \therefore \sin(30^\circ + \alpha) = 1 = \sin 90^\circ ∴ sin ( 3 0 ∘ + α ) = 1 = sin 9 0 ∘ 30 ∘ + α = 90 ∘ 30^\circ + \alpha = 90^\circ 3 0 ∘ + α = 9 0 ∘ α = 90 ∘ − 30 ∘ = 60 ∘ \alpha = 90^\circ - 30^\circ = 60^\circ α = 9 0 ∘ − 3 0 ∘ = 6 0 ∘
অতএব, α = 60 ∘ \alpha = 60^\circ α = 6 0 ∘ T 2 T_2 T 2
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, α = 30 ∘ \alpha = 30^\circ α = 3 0 ∘ W = 10 N W = 10\text{ N} W = 10 N
বলগুলোর মধ্যবর্তী কোণ:
T 2 T_2 T 2 W W W = 180 ∘ − α = 180 ∘ − 30 ∘ = 150 ∘ = 180^\circ - \alpha = 180^\circ - 30^\circ = 150^\circ = 18 0 ∘ − α = 18 0 ∘ − 3 0 ∘ = 15 0 ∘ T 1 T_1 T 1 W W W = 180 ∘ − 30 ∘ = 150 ∘ = 180^\circ - 30^\circ = 150^\circ = 18 0 ∘ − 3 0 ∘ = 15 0 ∘ T 1 T_1 T 1 T 2 T_2 T 2 = 30 ∘ + α = 30 ∘ + 30 ∘ = 60 ∘ = 30^\circ + \alpha = 30^\circ + 30^\circ = 60^\circ = 3 0 ∘ + α = 3 0 ∘ + 3 0 ∘ = 6 0 ∘
লামির সূত্র প্রয়োগ করে পাই:T 1 sin ( 180 ∘ − α ) = T 2 sin ( 180 ∘ − 30 ∘ ) = W sin ( 30 ∘ + α ) \frac{T_1}{\sin(180^\circ - \alpha)} = \frac{T_2}{\sin(180^\circ - 30^\circ)} = \frac{W}{\sin(30^\circ + \alpha)} s i n ( 18 0 ∘ − α ) T 1 = s i n ( 18 0 ∘ − 3 0 ∘ ) T 2 = s i n ( 3 0 ∘ + α ) W
বা, T 1 sin 150 ∘ = T 2 sin 150 ∘ = 10 sin 60 ∘ \frac{T_1}{\sin 150^\circ} = \frac{T_2}{\sin 150^\circ} = \frac{10}{\sin 60^\circ} s i n 15 0 ∘ T 1 = s i n 15 0 ∘ T 2 = s i n 6 0 ∘ 10
যেহেতু sin 150 ∘ = 1 2 \sin 150^\circ = \frac{1}{2} sin 15 0 ∘ = 2 1 sin 60 ∘ = 3 2 \sin 60^\circ = \frac{\sqrt{3}}{2} sin 6 0 ∘ = 2 3
T 1 = 10 × sin 150 ∘ sin 60 ∘ = 10 × 1 2 3 2 = 10 3 N ≈ 5.77 N T_1 = \frac{10 \times \sin 150^\circ}{\sin 60^\circ} = \frac{10 \times \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{10}{\sqrt{3}}\text{ N} \approx 5.77\text{ N} T 1 = s i n 6 0 ∘ 10 × s i n 15 0 ∘ = 2 3 10 × 2 1 = 3 10 N ≈ 5.77 N
একইভাবে,T 2 = 10 × sin 150 ∘ sin 60 ∘ = 10 3 N ≈ 5.77 N T_2 = \frac{10 \times \sin 150^\circ}{\sin 60^\circ} = \frac{10}{\sqrt{3}}\text{ N} \approx 5.77\text{ N} T 2 = s i n 6 0 ∘ 10 × s i n 15 0 ∘ = 3 10 N ≈ 5.77 N
উত্তর: T 1 = 10 3 N T_1 = \frac{10}{\sqrt{3}}\text{ N} T 1 = 3 10 N T 2 = 10 3 N T_2 = \frac{10}{\sqrt{3}}\text{ N} T 2 = 3 10 N
