Ok, I can solve Force Systems problem usually, but this onehas me thrown for a l

Question

Ok, I can solve Force Systems problem usually, but this onehas me thrown for a loop. The question says "At what angle must the 400-lb. forcebe applied in order that the resultant R of thetwo forces have a magnitude of 1000 lb? For this condition whatwill be the angle between R betweenR and the horizontal?" The book gives the answer of =51.3 , =18.19 I really need to understand this though, so words might evenbe more helpful than a solution. Do you sum the x and y components into Rx andRy and take an inverse trig function or something likethat? Thanks for the help.

Explanation / Answer

Rx = 400lb*cos + 700lb Ry = 400lb*sin R =[(Rx)2+(Ry)2 ] =[(400lb*cos+700lb)2 +(400lb*sin)2]    = [160000(lb)2*cos2 + 490000(lb)2+2*280000lb*cos +160000(lb)2sin2] = [160000(lb)2 + 490000(lb)2 + 560000lb2cos] = 1000lb => 650000 + 560000 cos = 1000000 cos = 35/28 = cos-1(35/56) = 51.3 Now Rx = 400lb*cos(51.3) + 700lb = 950.10lb Ry = 400lb*sin(51.3)=312.17 lb = tan-1(312.17/950.10) = 18.19

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