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I'm humbled friend. Thanks for such an encouraging comment, I really appreciate 🙏.

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Welcome friend. Thanks too for supporting this channel by watching our videos, I really appreciate. I wish you all the best in your studies 🙏.

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I've already done it here👉czcams.com/play/PLwyFA5ZBghfBbv9OSIEKjEniY8dMEobj-.html&si=7k23EVdZWiCd7fFw

I'm humbled friend. Thanks for such an encouraging comment, I really appreciate 🙏. I wish you all the best in your studies 🙏.

Welcome friend. Thanks too for your comment 🙏.

Welcome friend. Thanks too for such an encouraging comment, I really appreciate 🙏.

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Welcome friend. Thanks too for your comment.

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👍

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Welcome friend.

Welcome friend. Thanks too for supporting this channel by watching our videos, I really appreciate friend 🙏.

I'm humbled friend. Thanks for your comment, I will you all the best in your studies friend 🙏.

Welcome friend. I'm humbled. Thanks for such an encouraging comment, I really appreciate. I wish you all the best in your studies friend, be blessed 🙏.

Welcome friend. Ok, I'll work on that. Thanks for such an encouraging comment, I really appreciate friend. I wish you all the best in your studies friend, be blessed 🙏.

Welcome friend. Ok, I'll work on that. Thanks for such an encouraging comment, I really appreciate friend. I wish you all the best in your studies friend, be blessed 🙏.

Ok. Here is the working 👉 Question 1: Question 1: Let the distance of the 20N force from the pivot be x metres. Since a uniform metre rule has a total length of 1m or 100cm. It means the length of 80N force from the pivot will be (1-x) metres . Using the principle of moments; Sum of clockwise moment= Sum of anticlockwise moment about the pivot. That is: F1d1 = F2d2 ; 20*x = 80(1-x). On opening the brackets we get: 20x=80-80x. On collecting like terms, we get: 20x+80x=80 ; 100x=80. Dividing both sides by '100' we get x as: x=0.8m or 80cm Question 2: Question 2: From the principle of moments; Sum of clockwise moment = Sum of anticlockwise moment about the pivot, that is. F1d1+F2d2 = F3d3. Therefore: 30N*30/100 m + 20*(30+40)/100 m = R* 100/100 m. On simplifying we get: 9+14 = R hence R= 23N. Question 3: Question 3: Let the lengths X and Y be in metres. If the total length of the uniform bar = 50Cm = 50÷100 = 0.5 metre. Then, total length = X + Y = 0.5m. Making X to be the subject of the formula, we get X = 0.5 - Y ......... (1) From the principle of moments; Sum of clockwise moment = Sum of anticlockwise moment measured from the pivot F * d = F * d 3N * X = 5N * Y. ............ (2) Substituting equation (1) i.e X = 0.5 - Y in the equation (2) above we get; 3 * ( 0.5 - Y ) = 5 * Y On opening the brackets we have; 1.5 - 3Y = 5Y On collecting like terms we get 1.5 = 5Y + 3Y 1.5 = 8Y Dividing both sides by 8 we get 1.5 ÷ 8 = Y Therefore Y = 0.1875m or 0.1875*100 = 18.75Cm Substituting Y = 0.1875m in equation (1) i.e X = 0.5 - Y, we get X = 0.5 - 0.1875 Therefore X = 0.3125m or 0.3125 * 100 = 31.25Cm. Thanks for following my lessons.🙏

Ok. I'll work on that possibly in the coming school holiday. Thanks for your comment friend 🙏.

I'm humbled to hear this from you. Thanks for such an encouraging comment, I really appreciate friend. I'll keep uploading more for sure. Thanks for your comment 🙏.

Ok. I'm working on that. Welcome friend. Thanks too for your comment, I really appreciate 🙏.

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Wow, I'm humbled friend. I wish you all the best in your coming exams. Thanks for your comment friend 🙏. Be blessed.

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I'm very fine my friend. Welcome friend. Thanks too for supporting this channel by watching our videos, I really appreciate 🙏. Be blessed too 🙏.

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Hello too. Maybe you can give me an example of such a question so that I can see how best to approach it. Thanks for your comment friend 🙏.

Welcome friend. Thanks too for your support to this channel, I really appreciate 🙏.

Welcome friend. Thanks too for your support to this channel, I really appreciate 🙏.