Question

A ????=2.75m=2.75 kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a ????W=18.5FW=18.5 N horizontal force. Find the magnitude of the tension in the rope and the rope's angle from the vertical. The acceleration due to gravity is ????=9.81g=9.81 m/s2.

Answer 1

tension in the rope is 22.29 N

angle is 56.06 degree

Step-by-step explanation:

Given data

mass = 1.27 kg

horizontal force = 18.5 N

acceleration = 9.81 m/s2

to find out

magnitude and the rope angle

solution

we know object is in equilibrium so net force is zero here

so net force at x axis = 0 that mean F - Tsinθ = 0 , F = Tsinθ

and net force at y axis = 0 that is T cosθ - mg = 0 , T cosθ = mg

in equilibrium condition

F / mg = Tsinθ / T cosθ

18.5/ 1.27× 9.81 = sinθ / cosθ

so θ = tan^-1 (1.4859)

θ = 56.06 degree

angle is 56.06 degree

so F = Tsinθ

18.5 = T sin(56.06)

T = 18.5 / 0.8296

tension in the rope is 22.29 N

Answer 2

1) Tension in rope = 32.71 Newton's

2)
`\theta = 34.44^(o)`

Step-by-step explanation:

The system at equilibrium is shown in the attached diagram

For exulibrium in x dirextion we have

`\sum F_(x)=0\\\\\therefore 18.5-Tsin(\theta )=0\\\\Tsin(\theta )=18.5.....(i)\\Also\sum F_(y)=0\\\\\therefore Tcos(\theta )-mg=0\\\\Tcos(\theta )=mg....(ii)`

Dividing equation i and ii we get

`tan(\theta )=(18.5)/(mg)\\\\\therefore \theta =tan^(-1)((18.5)/(2.75* 9.81))\\\\\theta = 34.44^(o)`

Thus using this angle of inclination we get

`T=(18.5)/(sin(34.44))=32.71N`