Question

a block of mass M hangs in equilibrium. the rope which is fastened to the wall is horizontal and has a tension of 38N. the rope which is fastened to the ceiling has a tension of 59N and makes an angle theta with the ceiling. what is the angle

Answer 1

Answer:

50°

Explanation:

The tension in the rope fastened to the ceiling has horizontal component 59cosθ and its vertical component is 59sinθ. For equilibrium,

59cosθ = 38 N  (the tension in the rope attached to the wall) (1)and

59sinθ = Mg (the weight of the block of mass)  (2)

Dividing (2) by (1)

59sinθ/59cosθ = Mg/38

tanθ = Mg/38

θ = tan⁻¹(Mg/38)

Also squaring (1) and (2) and adding

(59cosθ)² = 38² N

(59sinθ)² = (Mg)²

(59cosθ)² + (59sinθ)² = 38² + (Mg)²

59²((cosθ)² + (sinθ)²) = 38² + (Mg)² . Since (cosθ)² + (sinθ)² = 1

59² = 38² + (Mg)²

Mg = √(59² - 38²) = √(3481 - 1444) = √2037 = 45.13 N

θ = tan⁻¹(Mg/38) = tan⁻¹(45.13/38) = tan⁻¹(1.1877) = 49.9° ≅ 50°

Answer 2

The angle of the theta along with the ceiling would be $50$°

Given that,

The tension of the rope linked to the wall $= 38N$

The tension of the rope linked to the ceiling $= 59N$

Since the tension of the rope linked to the wall is horizontal, it will be $59cos$θ while vertical element.

In order to establish the equilibrium,

$59cos$θ $= 38N$       ...(i)

$59sin$θ $= Mg$        ...(ii)

Now by solving through division of $(ii)$ by $(i)$,

$59sin$θ ÷ $59cos$θ $= Mg$ ÷ $38$

so,

$Tan$θ  $= Mg$ ÷ $38$

∵ θ = $Tan^-1 (Mg/38)$

To proceed further, we will square both  $(i)$ and  $(ii)$

($59cos$θ$)^2$ $= 38^2 N$

⇒ ($59sin$θ$)^2$ $= (Mg)^2$

⇒ ($59cos$θ$)^2$ + ($59sin$θ

by solving them through $(cos$θ$)^2$ $+ (sin$θ

$Mg = 45.13 N$

Now by putting the value of Mg in  $Tan^-1 (Mg/38)$, we get

θ $= 50$°

Thus,  $50$° would be the angle of the theta along with the ceiling.

Learn more about "Equilibrium" here:

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