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Asked by mayursamahajan | 09 Jun, 2021, 09:13: PM
Expert Answer
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 = { Ax2 + 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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