Physics, 07.04.2020 19:53 azariah7

The figure shows two wires that are tied to a 710 g mass that revolves in a horizontal circle at a constant speed of 7.5 m/s. (a) What is the tension in the upper wire? (b) What is the tension in the lower wire?(c) If the max tension in a wire can only be 30N. What would happen in this situation? What would be the new tension in order to maintain a speed of 7.5 m/s?

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Answer from: smith5845

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

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;

$F = \dfrac{m \cdot v^2}{r}$

r = The radius of rotation = The altitude of the triangle formed by the wires and the rod to which the wire are tied

$\therefore r = \dfrac{\sqrt{3} }{2} \times 1 \ m= \dfrac{\sqrt{3} }{2} \ m$

Which gives;

$F = \dfrac{0.71 \cdot 7.5^2}{ \dfrac{\sqrt{3} }{2} } = \dfrac{79.875\cdot \sqrt{3} }{3}$

Therefore;

$T_u \times sin(60^{\circ}) + T_l \times sin(60^{\circ}) = \dfrac{79.875\cdot \sqrt{3} }{3}$

Which gives;

$T_u + T_l = 2 \times \dfrac{79.875\cdot \sqrt{3} }{3} \times \dfrac{2 }{\sqrt{3} } = 53.25$

T_l = 53.25 - T_u

The vertical forces are presented as follow;

$T_u \times cos(60^{\circ}) = T_l \times cos(60^{\circ}) + m \cdot g$

Therefore;

$T_l \times cos(60^{\circ}) = T_u \times cos(60^{\circ}) - 0.71 \times 9.81$

$T_l = T_u - \dfrac{0.71 \times 9.81}{cos(60^{\circ})} = T_u -13.9302$

Which gives;

53.25- T_u = T_u - 13.9302

Solving gives;

The tension in the upper wire, T_u ≈ 33.5901 N

$(b) \ T_l \approx 33.5901 - \dfrac{0.71 \times 9.81}{cos(60^{\circ})} \approx 19.6599\ N$

The tension in the lower wire, T_l ≈ 19.6599 N

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;

$T_l = F = \dfrac{m \cdot v^2}{r}$

Which gives;

$T = F = \dfrac{0.71 \cdot 7.5^2}{ \dfrac{\sqrt{3} }{2} } = \dfrac{79.875\cdot \sqrt{3} }{3} \approx 42.12$

The new tension, T_l ≈ 42.12 N

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The figure shows two wires that are tied to a 710 g mass that revolves in a horizontal circle at a c

Answer from: tylermdons

(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

$F =\dfrac{M*v^2}{R}$

Since

$R = \sqrt{1.0m^2-0.5m^2} = \dfrac{\sqrt{3} }{2}$

m= 0.710kg ,

and v= 7.5m/s

the centripetal force is

$F =\dfrac{0.710(7.5)^2}{\dfrac{\sqrt{3} }{2} }$

F= 46.12N

The two wires are supporting this force; therefore,

$T_1 cos(\theta)+T_2 cos(\theta)=46.12N$ .

$(1).\:\: (T_u+T_d)cos(\theta)=46.12N$

Also, gravity is pulling down on the mass, and the wires must also support that; therefore,

$T_usin(\theta)= mg+T_dsin(\theta)$

$(2).\: \: (T_1-T_2)sin(\theta)= mg$

Now,

$\theta = sin^{-1}(\dfrac{0.5m}{1.0m} )= 30^o$ ;

therefore, equations (1) and (2) give

$(T_1+T_2)=\dfrac{46.12N }{cos(30^o)}$

$\boxed{(3). \:(T_1+T_2)=53.25N}$

and

$(T_1-T_2)= \dfrac{mg}{sin(\theta)}$

$\boxed{(4). \: T_1-T_2 = 13.916}$

Solving equations (3) and (4) gives

$\boxed{T_1= 33.59N}\\\\\boxed{T_2 = 19.67N}}$

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.

Answer from: Quest

the answer is b i think

Answer from: Quest

its a 5.5 n

glad that

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