a 6.00-kg ornament is held at rest by two light wires that form 30° angles with the vertical, as shown in the figure. an external force of magnitude f acts vertically downward on the ornament. the tension exerted by each of the two wires is denoted by t. a free-body diagram, showing the four forces that act on the box, is shown in the figure. if the magnitude of force f is 410 n, what is the magnitude of the tension t?

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Answer 1

The equilibrium condition allows finding the result for the tension of the cable that supports the block is:

The tension is:  T₁ = 270.66 N

Newton's second law gives a relationship between force, mass and the acceleration of bodies, when the acceleration is zero it is called the equilibrium condition.

A free-body diagram of the forces is shown in the attachment.

Let's write the equilibrium equation for the vertical axis.

        2 [tex]T_1_y[/tex] - f - W = 0

          2 [tex]T_1_y[/tex]  = W + f

Let's use trigonometry to find the tension, as the angle indicates that it is relative to the vertical.

        cos 30 = [tex]\frac{T_1_y}{T_1}[/tex]  

        [tex]T_1_y[/tex]  = T1 cos 30

     

We substitute.

        2T₁ cos 30 = W + f

         T₁ = m g + f / 2 cos 30

         T₁ = [tex]\frac{6 \ 9.8 + 410}{2 \ cos 30}[/tex]  

         T₁ = 270.66 N

In conclusion using the equilibrium condition we can find the result for the tension of the cable that supports en bloc is:

  The tension is:  270.66 N

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A 6.00-kg Ornament Is Held At Rest By Two Light Wires That Form 30 Angles With The Vertical, As Shown

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The maximum height reached by the ball during the upward motion  is 24.7 m.

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[tex]h = \frac{u^2}{2g} \\\\h = \frac{22^2}{2(9.8)} \\\\h = 24.7 \ m[/tex]

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