Find the sum of the given vectors 700N∠320∘+400N∠20∘ Rx=? Ry=? R=? θ=?

Question

Find the sum of the given vectors  700N∠320∘+400N∠20∘  Rx=?  Ry=?  R=?  θ=?

Jozlyn · Accepted Answer

Step 1  The given expression is, 700N∠230∘+400N∠20∘ The angle of vector is measured whith positive x-asis. Step 2 For 700 N, the angle made with positive x-axis is 230∘ Then angle made by 700 N with negative x-axis is 50∘ On resolving the vector, we get f1=700N[(−cos⁡(50∘))i^+(−sin⁡(50∘))j^] =−(697.34N)i^−(536.23N)j For 400 N, on resolving we get F2=400N[(cos⁡(20∘))i^+(sin⁡(20∘))j^] =(375.88N)i^+(136.81N)j^ Step 3 R=F1+F2 =[−(697.34N)i^−(536.23N)j^]+[(375.88N)i^+(136.81N)i^+(136.81N)j^] =[(−697.34+375.88)N]i^+[(−536.23+136.81)N]j^ =(−321.46N)i^+(−399.42N)j^ Hence, Rx=−321 Ry=−399 R=Rx2+Ry2 =(−321)2+(−399)2 =513 Step 4  tan⁡α=|Ry||Rx| α=tan−1,|Ry||Rx| =tan−1(399.42321.46) =51 Then angle made by resultant vector with positive x-axis is, θ=180∘+α =180∘+51.17∘

Find the sum of the given vectors displaystyle{700}{N}angle{320}^{circ}+{400}{N}angle{20}^{circ} Rx = ? Ry = ? R = ? theta = ?

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