$$\\\\\\text{From the diagram, it is clear that A and B are the two supports separated by 6m.}\\\\\\text{The weight of the plank (30kg) directly acts downwards at the centre of the plank about point C.}\\\\\\text{Suppose the man weighing 60 kg has reached the point (towards the right end) when the plank just begins to topple.}\\\\\\text{When the plank begins to topple, contact at support A is lost.}\\\\\\text{This helps us to equate the torques about support B.}\\\\\\text{Therefore, we have}\\\\{{F}_{C}}{{r}_{CB}}={{F}_{M}}{{r}_{BM}}\\\\\\text{Substituting the given value from the question in the above expression, we have}\\\\{{F}_{C}}{{r}_{CB}}={{F}_{M}}{{r}_{BM}} \\\\\\Rightarrow 30\\times 3=60\\times {{r}_{BM}} \\\\\\Rightarrow {{r}_{BM}}=\\dfrac{90}{60}\\\\\\therefore {{r}_{BM}}=1.5\\text{ m}\\\\\\text{Clearly, the man can do a distance of 1.5m from the nearer support without toppling the plank.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Torque Quiz 1","text":" A man and a boy carry a heavy weight by suspending it from a light pole of length $$\\text{6 m}$$. The ends of the pole rest on their shoulders and the pole is horizontal. At what distance from the boy should the weight be suspended so that the boy supports only half the weight supported by the man? ","comment":{"@type":"Comment","text":" Torques about an assumed point at which the weight needs to be suspended can be equated to determine the required distance."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3\\text{ m}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{3}\\text{ m}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2}{3}\\text{ m}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4\\text{ m}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\\\\\text{Let the heavy weight be denoted as w.}\\\\\\text{As provided in the question, the man needs to support a weight equal to}\\,\\dfrac{2w}{3}\\,\\text{whereas the boy needs to support a weight equal to}\\,\\dfrac{w}{3}.\\\\\\text{ Now, let us consider a diagram for the given situation.}\\\\\\text{From the diagram, we have label B for boy, M for man, distance OB as x and distance OM as (6-x).}\\\\\\text{Now, if we take the torques about O, we have}\\\\\\dfrac{w}{3}\\times x=\\dfrac{2w}{3}\\times (6-x) \\\\\\Rightarrow 3x=12 \\\\\\therefore x=4\\text{ m} \\\\\\text{Clearly, the weight should be suspended at a distance of 4 m from the boy to satisfy the given condition.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}