Answer :
The magnitude and direction of the force exerted by the second wire is 29.8 N at 238 degrees.
The sign hand attached to the wires is steady hence it is not moving anywhere. So, we can say that the total forces on the sign hand is zero.
Using vector addition, we can break down the force exerted by the first wire into its x- and y-components:
Fx = F₁cos(Ф₁)
= 31.0cos(122)
= -14.3 N (to the left)
Fy = F₁sin(Ф₁)
= 31.0sin(122)
= 26.5 N (upward)
The force exerted by the second wire must cancel out the horizontal component of the left wire and balance the vertical component, so:
F₂cos(Ф₂) = 14.3 N
F₂sin(Ф₂) = 26.5 N
Solving for F₂ and Ф₂, we get:
F₂ = 29.8 N
Ф₂ = 238 degrees
Therefore, the magnitude and direction of the force exerted by the second wire is 29.8 N at 238 degrees.
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