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The horizontal and vertical components of the force exerted on the beam at the wall



(A) horizontal component is 500N towards left and vertical component 75N downwards
(B) horizontal component is 500N towards right and vertical component 75N upwards
(C) horizontal component is 625N towards left and vertical component 150N upwards
(D) horizontal component is 625N towards right and vertical component 150N downwards

Answer
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Hint: The horizontal and vertical components of the force exerted on the beam at the wall can be determined by tension on the rope. To find the tension of the rope, the angle should be determined, then by using the angle value the tension is determined, then the horizontal and vertical force is determined by using the tension value.

Complete step by step solution:

From the diagram given in the question,
sinθ=35
By dividing the terms in the above equation, then
sinθ=0.6
By rearranging the terms in the above equation, then
θ=sin10.6
Then the angle value is, θ=37
Now, the tension equation for the given diagram is written as,
4T×sin37=(150×2)+(300×4)................(1)
At the point of the weight hangs, the vertical tension of the rope is given as 4Tsin37
The above equation (1) is the vertical force component equation, the RHS of the equation shows the equation of weight.
By multiplying the terms in the equation (1), then the above equation is written as,
4T×sin37=300+1200
The value of the term sin37=0.6, then the above equation is written as,
4T×0.6=300+1200
By multiplying the terms in the above equation, then the above equation is written as,
2.4×T=300+1200
By adding the terms in the above equation, then the above equation is written as,
2.4×T=1500
By rearranging the terms in the above equation, then the above equation is written as,
T=15002.4
By dividing the terms in the above equation, then the above equation is written as,
T=625N
The horizontal force is given by,
FH=Tcos37
By substituting the tension value in the above equation and also the value of the cos37=0.79863551, then the above equation is written as,
FH=625×0.79863551
By multiplying the terms in the above equation, then the above equation is written as,
FH=500N
Now, the vertical force is given by,
FV+Tsin37=150+300
By substituting the tension value in the above equation and also the value of the sin37=0.6, then the above equation is written as,
FV+625×0.6=150+300
By adding the terms in the above equation, then the above equation is written as,
FV+625×0.6=450
By rearranging the terms in the above equation, then the above equation is written as,
FV=450(625×0.6)
By dividing the terms in the above equation, then the above equation is written as,
FV=450375
By subtracting the terms in the above equation, then the above equation is written as,
FV=75N

Hence, the option (A) is the correct answer.

Note: The force of the horizontal component is acting on the left side because the tension of the rope pushes the rod towards the left side. And then the weight of the component pulls the rod downwards. The tension of the rope is used to determine the force.