A 1 125-N uniform boom at = 58.0° to the vertical is supported by a cable at an

Question

A 1 125-N uniform boom at = 58.0° to the vertical is supported by a cable at an angle = 32.0° to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight m = 2 300 N hangs from its top.

(a) Find the tension in the support cable (kN)

(b) Find the components of the reaction force exerted by the floor on the boom (Horizontal component magnitude & direction, vertical component magnitude and direction)

A 1 125-N uniform boom at phi = 56.0 degree to the vertical is supported by a cable at an angle theta = 32.0 degree to the horizontal as show in the figure below. The boom is the bottom, and an object of weight m = 2 300 N hangs from its top. Find the tension in the support cable. Find the components of the reaction force exerted by the floor on the boom.

Explanation / Answer

from equilibrium of forces :
Tcos = Rx

Tsin +Ry = 1125+2300

from torque equilibrium:

T(3l/4).sin(+) = (l/2).1125 .cos + l (2300)cos

T(3/4).sin(+) = (1/2).1125 .cos + (2300)cos

T(3/4).sin(32+58) = (1/2).1125 .cos58+ (2300)cos58

T(3/4) = (1/2).1125 .cos58+ (2300)cos58

T=2022 N

=2.022 kN

,...........

Rx = 1714.7 N = 1.714 kN

Ry = 6463

= 6.46 kN

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