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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 A

_{x}and A_{y}as shown in figure .At roller support B , We have reaction force B

_{y}as shown in figure.At Equilibrium , sum of Horizontal forces is zero. Hence we get ,

F sinα = A

_{x}+ P sinβ , Hence we get, A_{x}= F sinα - P sinβ = ( 320 -194 ) N = 126 NAt Equilibrium , sum of vertical forces is zero. Hence we get ,

A

_{y }+ B_{y}= 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 A

_{y}+ 10 B_{y}= 8 × P cosβ = 8 × 81 = 646 N m .................................(2)By solving eqn.(1) and (2) , we get , A

_{y}= 427 N and B_{y}= -106 NHence reaction forces at A :- A

_{x}= 126 N and A_{y}= 427 NResultant reaction at A = { A

_{x}^{2 }+ A_{y}^{2}}^{1/2}= { 126^{2}+ 427^{2}}^{1/2}= 445 NDirection of reaction force at A :-

if the direction of resultant makes angle θ with beam , then θ = tan

^{-1}( A_{y}/ A_{x}) = tan^{-1}( 427 / 126 ) = 73.5^{o}
Answered by Thiyagarajan K | 10 Jun, 2021, 01:09: AM

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