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# JEE Class main Answered

Asked by mayursamahajan | 09 Jun, 2021, 09:13: PM
Since weight of the beam is not given, let us assume weight of beam is negligible compare to the applied forces.

Applied forces F and P are resolved into horizontal and vertical components as shown in figure.

From the given slope of applied force F , Horizontal componenet of F and vertical componenet of F are given as

F sinα = 400 × 0.8 = 320 N

F cosα = 400 × 0.6 = 240 N

From the given slope of applied force P , Horizontal componenet of P and vertical componenet of P are given as

P sinβ = 210 × (12/13)  ≈ 194 N

P cosβ = 210 × (5/13)   81 N

At pin joint A , we have reaction forces Ax and Ay as shown in figure .

At roller support B , We have reaction force By as shown in figure.

At Equilibrium , sum of Horizontal forces is zero. Hence we get ,

F sinα  = Ax + P sinβ ,  Hence we get,  Ax = F sinα - P sinβ = ( 320 -194 ) N = 126 N

At Equilibrium , sum of vertical forces is zero. Hence we get ,

Ay + By = F cosα + P cosβ = ( 240 + 81 ) N = 321 N  .................................(1)

At equilibrium , sum of clockwise moment equals sum of counterclockwise moments.

Let us take moment of forces about left most edge of beam, then we get

4 Ay + 10 By = 8 × P cosβ = 8 × 81 = 646 N m .................................(2)

By solving eqn.(1) and (2) , we get , Ay = 427 N  and By = -106 N

Hence reaction forces at A :-  Ax = 126 N   and Ay = 427 N

Resultant reaction at A  = { Ax+ Ay2 }1/2 = { 1262 + 4272 }1/2 = 445 N

Direction of reaction force at A :-

if the direction of resultant makes angle θ with beam , then θ = tan-1 ( Ay / Ax ) = tan-1 ( 427 / 126 ) = 73.5o

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