Question

A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.60 m long and has a mass of 10.0 kg . The mass of the traffic light is 23.5 kg. 1.) Determine the tension in the horizontal massless cable CD. 2) Determine the vertical component of the force exerted by the pivot A on the aluminum pole. 3.) Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.

Answer 1

There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.

We need to be aware of the force in order to discover the solution.

How can I determine the tension in the CD's horizontal massless cable?

• The free body diagram of the masses must be drawn in order to determine the tension in the horizontal cable CD.
• To obtain the tension on CD in the free body diagram, let's balance all the vertical and horizontal forces.

•                        $TH-mg\frac{l}{2}cos\alpha -Mglcos\alpha =0\\T=\frac{glcos\alpha (\frac{m}{2}+M )}{h} \\T=446N$

where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,

How to calculate the force the pivot A exerted on aluminum's vertical and horizontal components?

• The overall force acting vertically is,

                      $F_V-mg-Mg=0\\F_v=328.2N$

• The overall force acting horizontally is,

                  $F_H=T=446N$

In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.

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Answer 2

1) The tension in the horizontal massless cable CD is, 446N

2) The vertical component of the force exerted by the pivot A on the aluminum pole 328.3N

3)The horizontal component of the force exerted by the pivot A on the aluminum pole is,446N

To find the answer, we need to know about the tension.

How to find the tension in the horizontal massless cable CD?

• To find the tension in horizontal cable CD ,we have to draw the free body diagram of the masses.
• Given that,

                    $l_{AB}=7.60m\\m_{AB}=10kg.\\M=23.5kg\\h=3.8 m\\\alpha =37degree.$

• In the free body diagram, let's balances all the vertical and horizontal forces to get the tension on CD.

                            $Th-mg\frac{l}{2}cos\alpha -Mglcos\alpha =0\\\\ T=\frac{mg\frac{l}{2}cos\alpha -Mglcos\alpha}{h}\\\\ T=\frac{glcos\alpha (\frac{m}{2}+M) }{y} \\\\T=\frac{9.8*7.6*cos37 (\frac{10}{2}+23.5) }{3.8} \\\\T=446 N$

How to find the vertical  and horizontal components of the force exerted by the pivot A on the aluminum pole?

• The net force along vertical direction can be written as,

                           $F_V-mg-Mg=0\\F_V= g(m+M)\\F_V=9.8*(10+23.5)=328.3 N$

• The net force along the horizontal direction is,

                           $F_H-T=0\\F_H=T=446N$

Thus, we can conclude that, the tension in the horizontal massless cable CD and the horizontal component of force exerted by the pivot A on the aluminum pole are same and is equal to 466N, and the vertical component force exerted by the pivot A on the aluminum pole is 328.3N.

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