Question

A 500N person stands 2.5m from a wall against which a horizontal beam is attached. The beam is 6m long and weighs 200N. A cable is attached to the free end of the beam and makes an angle of 45 to the horizontal and is attached to the wall.
1)Determine the magnitude of the tension of the cable.
2)Determine the reaction force that the wall exerts on the beam.

Answer 1

PART A)

Here by force balance along Y direction

$Tsin45 + F_y = 500 + 200$

Force balance along X direction

$Tcos45 = F_x$

now by torque balance

$Tsin45 (6) = 500 (2.5) + 200 (3)$

$T(4.24) = 1250 + 600$

$T = 436.3 N$

PART B)

now from above equations

$Tsin45 + F_y = 700$

$F_y = 391.5 N$

$F_x = Tcos45 = 308.5 N$

now net reaction force of wall is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = 498.4 N$