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19.A 20 kg sign is pulled by a horizontal force such that the single rope (originally vertical) holding the sign makes an angle of 21° with the vertical. Assuming the sign is motionless, findA) the magnitude of the tension in the rope andB) the magnitude of the horizontal force.

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Answer:

A)[tex]T=209.94N[/tex]

B) [tex]F=75.24N[/tex]

Explanation:

Using the free body diagram and according to Newton's first law, we have:

[tex]\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)[/tex]

A) Solving (1) for T:

[tex]T=\frac{mg}{cos(21^\circ)}\\T=\frac{20kg(9.8\frac{m}{s^2})}{cos(21^\circ)}\\T=209.94N[/tex]

B) Solving (2) for F:

[tex]F=Tsin(21^\circ)\\F=(209.94N)sin(21^\circ)\\F=75.24N[/tex]

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