Answer:
tension in the rope is 22.29 N
angle is 56.06 degree
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:
1) Tension in rope = 32.71 Newton's
2)[tex]\theta = 34.44^{o}[/tex]
Explanation:
The system at equilibrium is shown in the attached diagram
For exulibrium in x dirextion we have
[tex]\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)[/tex]
Dividing equation i and ii we get
[tex]tan(\theta )=\frac{18.5}{mg}\\\\\therefore \theta =tan^{-1}(\frac{18.5}{2.75\times 9.81})\\\\\theta = 34.44^{o}[/tex]
Thus using this angle of inclination we get
[tex]T=\frac{18.5}{sin(34.44)}=32.71N[/tex]
