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solniwko [45]
3 years ago
6

A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cab

le was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable
Physics
1 answer:
Lorico [155]3 years ago
4 0

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        F_{y} + F_{y} - W =0

        2F_{y} = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = \frac{ 800}{ 2sin 2.29}

        T = 10010 N

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A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp
AlekseyPX

Answer:

(A) 2.4 N-m

(B) 0.035kgm^2

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque \tau =Fr, here F is force and r is distance

So \tau =20\times 0.12=2.4Nm

(B) Moment of inertia is equal to I=\frac{1}{2}mr^2

So I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2

Torque is equal to \tau =I\alpha

So angular acceleration \alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2

(C) As the disk starts from rest

So initial angular speed \omega _{0}=0rad/sec

Time t = 4.6 sec

From first equation of motion we know that \omega =\omega _0+\alpha t

So \omega =0+68.571\times  4.6=315.426rad/sec

(D) Kinetic energy is equal to KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J

(E) From second equation of motion

\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad

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3 years ago
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