Answer:
(a). Upper wire = 33.59 N
(b). downward wire = 19.67 N
(c). The upper wire would break.
Explanation:
The centripetal force that keeps the mass moving in a circular path is directed towards the center and has the magnitude
[tex]F =\dfrac{M*v^2}{R}[/tex]
Since
[tex]R = \sqrt{1.0m^2-0.5m^2} = \dfrac{\sqrt{3} }{2}[/tex]
[tex]m= 0.710kg[/tex],
and [tex]v= 7.5m/s[/tex]
the centripetal force is
[tex]F =\dfrac{0.710(7.5)^2}{\dfrac{\sqrt{3} }{2} }[/tex]
[tex]F= 46.12N[/tex]
The two wires are supporting this force; therefore,
[tex]T_1 cos(\theta)+T_2 cos(\theta)=46.12N[/tex].
[tex](1).\:\: (T_u+T_d)cos(\theta)=46.12N[/tex]
Also, gravity is pulling down on the mass, and the wires must also support that; therefore,
[tex]T_usin(\theta)= mg+T_dsin(\theta)[/tex]
[tex](2).\: \: (T_1-T_2)sin(\theta)= mg[/tex]
Now,
[tex]\theta = sin^{-1}(\dfrac{0.5m}{1.0m} )= 30^o[/tex];
therefore, equations (1) and (2) give
[tex](T_1+T_2)=\dfrac{46.12N }{cos(30^o)}[/tex]
[tex]\boxed{(3). \:(T_1+T_2)=53.25N}[/tex]
and
[tex](T_1-T_2)= \dfrac{mg}{sin(\theta)}[/tex]
[tex]\boxed{(4). \: T_1-T_2 = 13.916}[/tex]
Solving equations (3) and (4) gives
[tex]\boxed{T_1= 33.59N}\\\\\boxed{T_2 = 19.67N}}[/tex]
Thus, the tension in the upper wire is 33.59 N and in the bottom wire 19.67 N.
(c).
If the maximum tension in a wire can only be 30 N, then the upper wire would break because we see that its tension of 33.59 N exceeds 30 N.
The weight of the mass is included to the tension of the string when two
strings are attached but not included when using only one.
(a) The tension in the upper wire is approximately 33.5901 N
(b) The tension in the lower wire is approximately 19.6599 N
(c) The new tension using only one string in horizontal circular motion is approximately 42.12 N
Reasons:
The given parameter are;
Mass attached to the wires, m = 710 g = 0.71 kg
Speed or rotation of the wires, v = 7.5 m/s
Length of wires = 1.0 m
Distance apart of the base of the wires = 1.0 m
(a) The tension in the upper wire is given as follows;
Horizontal force in the wire = The centripetal force of thee revolving mass
The centripetal force of thee revolving mass is given by the formula;
[tex]F = \dfrac{m \cdot v^2}{r}[/tex]
r = The radius of rotation = The altitude of the triangle formed by the wires and the rod to which the wire are tied
[tex]\therefore r = \dfrac{\sqrt{3} }{2} \times 1 \ m= \dfrac{\sqrt{3} }{2} \ m[/tex]
Which gives;
[tex]F = \dfrac{0.71 \cdot 7.5^2}{ \dfrac{\sqrt{3} }{2} } = \dfrac{79.875\cdot \sqrt{3} }{3}[/tex]
Therefore;
[tex]T_u \times sin(60^{\circ}) + T_l \times sin(60^{\circ}) = \dfrac{79.875\cdot \sqrt{3} }{3}[/tex]
Which gives;
[tex]T_u + T_l = 2 \times \dfrac{79.875\cdot \sqrt{3} }{3} \times \dfrac{2 }{\sqrt{3} } = 53.25[/tex]
[tex]T_l = 53.25 - T_u[/tex]
The vertical forces are presented as follow;
[tex]T_u \times cos(60^{\circ}) = T_l \times cos(60^{\circ}) + m \cdot g[/tex]
Therefore;
[tex]T_l \times cos(60^{\circ}) = T_u \times cos(60^{\circ}) - 0.71 \times 9.81[/tex]
[tex]T_l = T_u - \dfrac{0.71 \times 9.81}{cos(60^{\circ})} = T_u -13.9302[/tex]
Which gives;
[tex]53.25- T_u = T_u - 13.9302[/tex]
Solving gives;
The tension in the upper wire, [tex]T_u[/tex] ≈ 33.5901 N
[tex](b) \ T_l \approx 33.5901 - \dfrac{0.71 \times 9.81}{cos(60^{\circ})} \approx 19.6599\ N[/tex]
The tension in the lower wire, [tex]T_l[/tex] ≈ 19.6599 N
(c) If the max tension in a wire can only be 30 N, the upper wire will break
first followed by the lower wire.
In order to maintain the speed of 7.5 m/s, the new tension should be above
33.6 N, for the two wires and, when only one wire, is attached, we have;
[tex]T_l = F = \dfrac{m \cdot v^2}{r}[/tex]
Which gives;
[tex]T = F = \dfrac{0.71 \cdot 7.5^2}{ \dfrac{\sqrt{3} }{2} } = \dfrac{79.875\cdot \sqrt{3} }{3} \approx 42.12[/tex]
The new tension, [tex]T_l[/tex] ≈ 42.12 N
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