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InterviewSolution
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Two bodies A and B of mass 5 kg and 10 kg respectively in contact with eachother rest on a table against a rigid partition The coefficient of friction between the bodies and the table is 0.15 A force of 200 N is applied horizontally at A Calculate (i) reaction of the partition (ii) action reaction forces between A and B What happens when the partition is removed ? Does the anwer to (ii) change when the bodies are in motion ? Ignore difference between `mu_(s)` and `mu_(k)` . |
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Answer» Here , mass of body A, `m_(1) = 5 kg` mass of body B, `m_(2) = 10 kg` coefficient of friction between the bodies and the table ` mu = 0.15 ` Horizontal force applied on A , `F = 200 N ` (i) Force of limiting friction acting to the left `f = mu (m_(1) + m_(2) g` `= 0. 15 (5 + 10 ) xx 9.8 = 22 .05 N` Net force exerted on the partition (to the right) `F = F - f = 200 - 22.05 = 177.95 N ` Reaction of partition `= 177 . 95 N` to the left Force of limiting friction acting on the body A `f_(1) = mu m_(1) g = 0 .15 xx 5 xx 9.8 = 7 . 35 N` Net force exerted by body A on body B `F' = F - f_(1) = 200 - 7 .35 = 192 . 65 N` This force is to the right Reaction of body B on body A = 192 . 65 N to the left Acceleration produced in the system `a = (F)/(m_(1) + m_(2)) = (177 . 95)/(5 + 10) = 11 .86 m//s^(2)` Force producing motion in body A `F_(1) = m_(1) a = 5 xx 11.86 = 59.3 N` Net force exerted by body A on B when partition is removed ` = F ' - F_(1) = 192.65 - 59.3 = 133.35 N` Hence reaction of body B on body A when partition is removed `= 133 . 35 N` Thus answers to (ii) do change . |
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