Question

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.

Answer 1

Answer:

A)$T=209.94N$

B) $F=75.24N$

Explanation:

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

$\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)$

A) Solving (1) for T:

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

B) Solving (2) for F:

$F=Tsin(21^\circ)\\F=(209.94N)sin(21^\circ)\\F=75.24N$