Ask your own question, for FREE!
Physics 23 Online
OpenStudy (callisto):

Can anyway help check my explanation to the question? for (a) , |R3| = |R2| R2 sinθ = 30 => R2 > 30 So (a) is correct For (b) R5 = 120 + R3cosθ = 120 + 30cosθ > 150 So (b) is incorrect For (c) since R2 and R3 are action and reaction pair, R2sinθ = R3 sinθ Consider the whole system, R1 + R3 = R2+R4 So R1 = R4 for (d), Hmm.. don't know how to explain... But i know they form an action and reaction pair...

OpenStudy (callisto):

OpenStudy (anonymous):

wat is ur answer?

OpenStudy (callisto):

if you're asking 'is', then it is B, but for 'was', it was 'A'

OpenStudy (anonymous):

u changed ur mind to b but why?

OpenStudy (callisto):

i made up some explanations for me to understand :(

OpenStudy (anonymous):

the incorrect option is a) as the weight of body is rested at two places rather than one so the weight gets distributed so in no way can the r3 be greater than 30 also note that r3=r2

OpenStudy (callisto):

but then the answer is B .. very confusing..

OpenStudy (anonymous):

if that were true also then how can r3 be greater than 30 that is near impossible

OpenStudy (callisto):

R2 sinθ = 30 => R2 > 30

OpenStudy (callisto):

that's the only possible explanation from me..

OpenStudy (callisto):

R2 is resolved into 2 components..

OpenStudy (anonymous):

ok sin@ of some angle reduces its measur remember maximum value of sine is 1

OpenStudy (callisto):

that would be for 90 degrees only... obviously it is not that case, 0<sinθ<1

OpenStudy (callisto):

So when a number is divided by a decimal, it becomes larger..

OpenStudy (anonymous):

i am not able to explain in mathematical terms but think abt it when a body rests at more than one location the weight is shared equally right

OpenStudy (callisto):

|dw:1332691872284:dw| since it is in eqm, R2 cosθ = W R2 cos θ = 30 R2 = 30/cosθ for 0<θ <1

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!