NEET Class neet Answered
Three forces 9,12,and 15 acting at a point are in equilibrium. The angle between 9N And 12N is
Asked by vivekkumarsoni268 | 25 Aug, 2021, 03:11: PM
Expert Answer
Figure shows the three forces 9N , 12N and 15N that are represented by OA , OB, and OC in equilibrium.
Let OD be the line along the force 15N . Angle α and β are made by 9N and 12 N with line OD respectively as shown in figure.
If we resolve the 9N and 12 N forces along OD , then at equilibrium we have
9 cosα + 12 cosβ = 15 or 9 cosα = 15 - 12 cosβ .........................(1)
If we resolve the 9N and 12 N forces along line perpendicular to OD , then at equilibrium we have
9 sinα= 12 sinβ ...............................(2)
By squaring and addin eqn.(1) and (2) , we get
81 = 144 + 225 - 360 cosβ
By simplifying above expression , we get cosβ = 4/5 , Hence sinβ = 3/5
From eqn.(2) , we get , sinα = (12/9) sinβ = (12/9) (3/5) = 4/5
Angle between 9N and 12N is ( α + β )
sinα = 4/5 = 0.8 , hence α = sin-1(0.8) = 53o
sinβ = 3/5 = 0.6 , hence β = sin-1(0.6) = 37o
Angle between 9N and 12N = ( α + β ) = 90o
( Though answer is not matching with the given options , this is the correct answer )
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