Respuesta :
Explanation:
a)
Sum of moments = 0 (Equilibrium)
T . cos (Q)*L = m*g*L/2
[tex]cos Q = \frac{\sqrt{(2.5^2 - L^2) } }{2.5}[/tex]
[tex]T*\frac{\sqrt{(2.5^2 - L^2) } }{2.5} * L = 0.1 *9.81*L/2[/tex]
[tex]T = \frac{2.4525}{\sqrt{(2.5)^2 - L^2} }[/tex]
b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!
c)
[tex]T = \frac{1.962}{\sqrt{(2)^2 - L^2} }[/tex]
The rotational equilibrium condition allows finding the responses for the tension of the rope are:
A) T = 0.53 N
B) If the rope shortens the tension increase..
C) T = 0.57 N
Newton's second law for rotational motion gives a relationship between the torque, the moment of inertia and the angular acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition.
Σ τ = 0
Torque is defined as the vector product of the force and the distance, its modulus is:
τ = F rsin θ
where τ is the torque, F the force, r the distance and tea the angle between the force and the distance, it is a product (r sin θ ) it is called the perpendicular distance or arm.
In the attached we have a free body diagram of the system, let's apply the equilibrium condition,
Let's use trigonometry to descompose the force.
cos θ = [tex]\frac{T_x}{T}[/tex]
sin θ = [tex]\frac{T_y}{T}[/tex]
Tₓ = T cos θ
[tex]T_y[/tex] = T sin θ
They indicate that the length of the bar is x = 1 m and the length of the cable is L = 2.5 m, let's find the angle
cos θ = [tex]\frac{x}{L}[/tex]
θ = cos⁻¹ [tex]\frac{x}{L}[/tex]
θ = cos⁻¹ [tex]\frac{1}{2.5}[/tex]
θ = 66.4º
A) let's set our pivot point at the junction with the wall and the anti-clockwise direction of rotation is positive.
W [tex]\frac{L}{2}[/tex] - Ty L = 0
W [tex]\frac{L}{2}[/tex] = (T sin 66.4) L
[tex]T = \frac{mg}{s sin 66.4} \\ \\T = \frac{0.1 \ 9.8}{2 \ sin66.4}[/tex]
T = 0.53 N
B) how the tension changes as the length of the string changes.
T = [tex]\frac{mg}{2 sin \theta}[/tex]
we can see that the change of the tension occurs by changing the value of the sine function.
sin θ = [tex]\frac{y}{L_o}[/tex]
Let's use the Pythagorean theorem to find the opposite leg.
L² = x² + y²
y = [tex]\sqrt{L^2 - x^2 }[/tex] = [tex]L \ \sqrt {1^2 + (\frac{x}{L})^2 }[/tex]
Let's substitute.
sin θ = [tex]\sqrt{1 - \frac{x^2}{L^2} }[/tex]
if we use a binomial expansion.
[tex](1 - a) ^{0.5} = 1 - \frac{1}{2} a + ...[/tex]
Let's substitute.
sin θ = 1 - [tex]\frac{1}{2} \ \frac{x}{L}[/tex]
We can see that when the value of the length decreases the value of the sine decreases and this term is in the denominator of the expression of the tension, therefore the tension must increase.
C) the length of the rope is L = 2 m
sin θ = [tex]\sqrt{1 - (\frac{1}{2})^2 }[/tex]
sin θ = 0.866
T = [tex]\frac{mg}{2sin \theta}[/tex]
T = [tex]\frac{0.1 \ 9.8}{2 \ 0.866}[/tex]
T =0.57 N
In conclusion, using the rotational equilibrium condition we can find the answers for the tension of the rope are:
A) T = 0.53 N
B) If the rope shortens the tension increase.
C) T = 0.57 N
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