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forces of 3, 4, 5, 6 and 25 act respective along the sides AB, BC, CD and DA along the diagonal AC of the square ABCD find their resultant. |
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Answer» Explanation:ABCD is a rectangle in which AB=4 m and BC=3 m Then, tanθ= AB BC = 4 3 The forces 3,P,5,10 and Q NEWTONS have the resultant R Newton as SHOWN in the figure. Rcosθ+5−3 =Qcosθ+2....(i) and Rsinθ=P+10−Qsinθ....(ii) and Q ABsinθ+5.BC=10.AB or Q 4.(3/5)+5×3=10×4(∵sinθ= 5 3 and cosθ= 5 4 ) ⇒ 5 12 Q=40−15=25 ⇒Q= 12 125 newton Then, find R from Eq. (i) and P from Eq. (ii). 5 4R = 12 125 × 5 4 +2 ⇒4R= 3 125 +10= 3 155 ⇒R= 12 155 and 12 155 × 5 3 =P+10− 12 125 × 5 3 4 155 =5P+50− 4 125 ⇒5P= 4 155 + 4 125 −50 ⇒5P= 4 280 −50=70−50=20 ⇒P=4. |
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