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

A)
`T=209.94N`

B)
`F=75.24N`

Step-by-step 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=(mg)/(cos(21^\circ))\\T=(20kg(9.8(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`