The resultant of the two forces #3N# and#2N# at an angle #theta# is doubled when first force in increased to #6N#. Find #theta#?

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Answer 1 · 2016-03-07T05:29:30

#theta=(2pi)/3# or #120^@#

Formula for resultant force of two forces #P# and #Q# is given by #R=sqrt(P^2+Q^2+2PxxQxxcostheta)#.

When two forces #3N# and #2N# are at an angle #theta#, value of the first resultant is #R_1# is given by

#R_1=Nsqrt(3^2+2^2+2xx3xx2xxcostheta)=sqrt(13+12costheta)#

When first force is increased to #6N#, the resultant force #R_2# is given by

#R_2=Nsqrt(6^2+2^2+2xx6xx2xxcostheta)=sqrt(40+24costheta)#

As #R_2# is doubled

#Nsqrt(40+24costheta)=2xxNsqrt(13+12costheta)# or

#40+24costheta=4xx(13+12costheta)=52+48costheta#

Hence, #24costheta=-12# or #costheta=-1/2#

and #theta=(2pi)/3# or #120^@#