A 1127.0N uniform boom of length L is supported by a cable. The boom is pivoted

Question

A 1127.0N uniform boom of length L is supported by a cable. The boom is pivoted at the bottom, the cable is attached a distace 3/4L from the pivot, and a 2731.0N weight hangs from the booms top. The force FT applied by the supporting cable is 2392.41N. The angle of the cable attached to the boom from the wall is 33 degrees. And the angle of the boom to the ground is 57 degrees.

1) Find the horizontal component of the reaction force on the bottom of the boom.

2) Find the vertical component of the reaction force on the bottom of the boom.

Explanation / Answer

Please go through the answer .....i have answered before....so go through it .

Q) A 1130.0N uniform boom of length L is supported by a cable. The boom is pivoted at the bottom, the cable is attached at a distance 3/4L from the pivot, and a 4818.0N weight hangs from the booms top.

Find the force Ft applied by the supporting cable.
Find the horizontal and vertical componenta of the reaction force on the bottom of the boom.


21 degree triangle the hypotenuse=Ft
another straight line that angles at 69 degrees touches the end of the triangle and has the 4818N boom hanging off of it

ANS ) -------------------------------------------------------------->>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

There are a number of forces acting here:
Force of gravity on boom's center (Fg1 down).
Force of gravity on weight (Fg2 down).
Force of tension in cable (Ft and 21 degrees).
Hinge force by pivot (Fh and angle x).

Sum of vertical forces = sum of vertical forces
Fy = Fy
ma = Ft sin 21 + Fh sin x - Fg1 - Fg2
0 = Ft sin 21 + Fh sin x - 1130 - 4818
5948 = Ft sin 21 + Fh sin x
Fh sin x = 5948 - Ft sin 21

Sum of horizontal forces = sum of horizontal forces
Fx = Fx
ma = Ft cos 21 - Fh cos x
0 = Ft cos 21 - Fh cos x
Fh cos x = Ft cos 21

Sum of torques = sum of torques
Fr = (Ft * sin 21 * 3/4 L) - (Fg1 * L/2) - (Fg2 * L) + (Fh * 0)
0 = .75 sin 21 L Ft - .5 (1130) L - 4818 L
0 = .269 L Ft - 5383 L
0 = L(.269 Ft - 5383)
0 = .269 Ft - 5383
5383 = .269 Ft
Ft = 20028 N (#1)

Fh cos x = Ft cos 21
Fh cos x = 20028 cos 21
Fh cos x = 18698 N (#2a)

Fh sin x = 5948 - Ft sin 21
Fh sin x = 5948 - 20028 sin 21
Fh sin x = -1229.3 N (#2b) ANS )

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