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sphere of mass M is held between 2 smooth inclined planes for Sin 37 is equal to 3 by 5 the normal reaction of the wall 2 to is equal to
Mg=2Rsin37R=Mg/2sin37R=Mg / 2*3/5 ( sin 37=3/5 is given) therefore , R=5Mg/6this sphere is lying between 2 inclined planes so there are 2 normal reactions on the sphere by the plane. if we resolve these reactions the vertical forces will be Rsin37*2 this system is under equllibrium so verical forces 2Rsin37 is equal to weight of the sphere Mg (here assume normal reaction= R)
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