A 588 N uniform boom is supported by a cable. The boom is pivoted at the bottom and a 1770 N object hangs?

Question

from its end. The boom has a length L= 24m and is at an angle of 48deg above the horizontal. A support cable is attached(at a 90deg angle) to the boom at a distance of 0.66 L from the foot of the boom and its tension is perpendicular to the boom.
a)find the tension T in the support cable in units N.
b)find the horizontal component of the reaction force on the boom by the floor in units N.
c)find the vertical components of the reaction force on the boom by the floor in units N.

Answer 1

First,
compute the components of force perpendicular to the boom due to the weight of the boom and the object:
Bp=cos(48)*588
Op=cos(48)*1770

the tension in the support cable can be computed by summing the torques around the pivot point:

TC+TBp+TOp=0
TC=tension*.66*24
TBp=-Bp*.5*24
TOp=-Op*24
Tension=(Bp*.5+Op)/.66
=2093N

The vertical component of the tension is
cos(48)*2093
The horizontal component is
sin(48)*2093
=1555N
which is equal to the horizontal reaction force in the floor

Since we know the components of force on the floor parallel to the beam as
Bh=sin(48)*588
Oh=sin(48)*1770

then v^2+h^2=parallel^2
v=sqrt((sin(48)*588+sin(48)*1770)^2-1555^2)
=807N


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Answer 2

A 50 gm Rosogolla has 93 calories. Cheers!