As shown in figure 10 N force is applied at two ends of a rod. Calculate tensile stress and searing stress for section PR, Area of cross section PQ is 10 cm^(2).

As shown in figure 10 N force is applied at two ends of a rod. Calcula

1.	As shown in figure 10 N force is applied at two ends of a rod. Calculate tensile stress and searing stress for section PR, Area of cross section PQ is 10 cm^(2).
Answer» Solution :Suppose cross section PQ having area A and PR having A. `therefore A. =(A)/(cos 30 ^(@)) = (10 xx 10 ^(-4))/( sqrt 3 // 2 ) ` `therefore A. = ( 2 xx 10 ^(-3))/( sqrt3) m ^(2) ""...(1)` Here, angle between force `vecF` and ara of A. cross-section PR is `30^(@).` Force `vecF` has two component `(i) F cos theta,` PARALLEL component to A. which provides tension `therefore F_(l) = F cos 30^(@)` `=10 xx (sqrt3)/( 2)` `= 5 sqrt3 N ""...(2)` (II) `F sin theta,` perpendicular component to A., it is tangential `therefore F _(t) = F sin 30^(@)` `=10 xx 1/2` `therefore F_(t) = 5 N ""...(3)` Tensile stress `SIGMA _(1) = ("Tensile force "(F _(t)))/(A.)` `= ( 5 sqrt3 xxsqrt3)/(2xx10 ^(-3)` (From equation (1) and (2)) `therefore sigma _(l ) = 7.5 xx 10 ^(+3) N //m ^(2)` Tangential (shear) stress ` sigma _(t) = ("Tangential Force" (F _(t)))/(A.)` `= ( 5 xx sqrt3)/( 2 xx 10 ^(3)) ` (From equation (1) and (3)) `therefore sigma _(t) = ( 5 sqrt3 xx 10 ^(3))/( 2) = 4.33 xx 10 ^(3) N//m ^(2)`