Please answer this question 

Asked by mayursamahajan | 9th 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  = { 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
 

Answered by Thiyagarajan K | 10th Jun, 2021, 01:09: AM