Answer :
Answer:20.97g N,32.63g N
Explanation:
We consider the forces at the knot.
The vertical forces are
[tex]T_{2}Sin(50^{0}})[/tex] is the vertical component of tension [tex]T_{2}[/tex] at the knot.
[tex]-25g[/tex] is the weight of the mass [tex]25Kg[/tex] acting downwards.
The horizontal forces are
[tex]-T_{1}[/tex] is the tension in the rope acting left.
[tex]T_{2}Cos(50^{0})[/tex] is the horizontal component of tension [tex]T_{2}[/tex] acting towards right.
Since the knot has no mass,it is always in equilibrium.
So,the sum of forces acting on it will be zero.
Balancing vertical forces gives,
[tex]T_{2}Sin(50^{0}})[/tex][tex]-25g=0[/tex]
[tex]T_{2}[/tex]=[tex]32.63gN[/tex]
Balancing horizontal forces gives,
[tex]T_{2}Cos(50^{0})[/tex][tex]-T_{1}=0[/tex]
[tex]T_{1}=32.63g\times Cos(50^{0})=20.97gN[/tex]
Answer:
The tension in the rope, T₁ = 205.6 N
The tension in the rope, T₂ = 319.8 N
Explanation:
Given data,
The suspended mass, m = 25 kg
The angle with the horizontal, Ф = 50°
Let the vertical is the y-axis and horizontal is the x-axis,
Then, only T₂ has components in both x and y.
For equilibrium to exist, the magnitude of the component of T₂ in x, T₂ₓ, is equal to T₁,
The T₂y is equal to the force 'F' due to the 25 kg mass
The force F on mass, F = 245 N = T₂y.
T₂y = T₂ sin 50°
Therefore,
T₂ = T₂y / sin 50°
= 245 N / 0.766
= 319.8 N
Hence, the tension in the rope, T₂ = 319.8 N
T₁ = T₂ₓ
= T₂ cos 50°
= 319 x 0.6428
= 205.6 N
Hence, the tension in the rope, T₁ = 205.6 N