Figure 3 shows a stationary metal block hanging from the middle of a stretched wire which is suspended from a horizontal beam. The tension in each half of the wire is 15 N. (a) Calculate for the wire at A, (i) the resultant horizontal component of the tension forces, (ii) the resultant vertical component of the tension forces. Can someone explain why the first one is 0?

(i) The horizontal component of the tension in the right wire points right, and the horizontal component of the tension in the left wire points left. Since these components are equal and opposite, the resultant horizontal force is 0.

∑Fₓ = 15 N cos 20° − 15 N cos 20°

∑Fₓ = 0 N

(ii) ∑Fᵧ = 15 N sin 20° + 15 N sin 20°

∑Fᵧ = 10.26 N

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abidemiokin abidemiokin · Answer 2 · 2021-09-30T15:34:04+02:00

i) The resultant horizontal component of the tension forces is 0N

ii) The resultant vertical component of the tension forces is 10.26N

iii) The sum of force along the horizontal is zero because the force has the same magnitude and moving in the opposite direction.

i) From the given figure, the horizontal component of the force is given as;

[tex]F_x = + 15 cos 20^0 + (-15cos20^0)\\F_x = + 15 cos 20^0 -15cos20^0\\F_x =0[/tex]

The sum of force along the horizontal is zero because the force has the same magnitude and moving in the opposite direction.

ii) The vertical component of the force is expressed as

[tex]F_y = + 15 sin 20^0 + 15sin20^0)\\F_y = 2(15sin20^0)\\F_y=2(5.13)\\F_y=10.26N[/tex]

Hence the resultant vertical component of the tension forces is 10.26N

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