Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is exerting an equal amount of force how much force does each wire exert?



Answer :

1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle)  to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
 
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos 

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. :  sin(40) x '?' = 4905
-divide both sides by sin(40):  '?' = 4905/ sin(40)
--> Solve.

The force each wire exert will be "390.62 N".

The given values are:

Weight,

  • mg = 500 N

Angle,

  • θ = 100°

As we know,

The sum of vertical forces = 0

then,

→ [tex]\Sigma Fy = 0[/tex]

Now,

→ [tex]F Cos (50) +F Cos (50) = mg[/tex]

By substituting the value, we get

→                   [tex]2F Cos (50) = 500[/tex]

→                                 [tex]F = \frac{500}{2 Cos (50)}[/tex]

→                                     [tex]= \frac{500}{2(0.64)}[/tex]

→                                     [tex]= 390.62 \ N[/tex]  

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