An angler hangs a 4.50-kg fish from a vertical steel wire...

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 $\times$ 10$^{-3}$ cm$^2$ in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force $\overrightarrow{F}$ at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force $\overrightarrow{F}$, (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). $\textbf{Torques and Tug-of-War.}$ In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0$^\circ$ to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is ${T_1} =$ 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, ${T_2}$ is not equal to ${T_1}$. His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 $\times$ 10$^{-3}$ cm$^2$ in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force $\overrightarrow{F}$ at the lower end of the wire, pulling it very slowly downward by 0.500 mm from
its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force $\overrightarrow{F}$, (d) the work done by the force the wire exerts on the fish; and
(e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). $\textbf{Torques and Tug-of-War.}$ In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0$^\circ$ to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is ${T_1} =$ 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, ${T_2}$ is not equal to ${T_1}$. His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

Transcript

00:03 Hello.

00:05 Now we will be working on the problem 11 .91 of young's book.

00:10 Here we have a wire hanging from a support and a fish hanging from the wire at the bottom.

00:19 My given values are the mass of the wire, the unstretched length of the wire, the cross -sectional area of the wire.

00:28 I'm sorry, this is the mass of the fish, this is the unstretched length of the wire.

00:33 And this one is the cross -sectional area of the wire and g is my acceleration due to gravity.

00:42 So we have to assume here that the wire is completely massless.

00:47 Now you can see that the wire is in tension here because of the fish hanging from it.

00:53 Let's draw the forces acting on the wire and on the fish.

00:57 I have already drawn them here.

00:59 So this is the free body diagram of the wire.

01:02 So we have tension forces acting on the wire because of the fish hanging from it.

01:08 And this is the free body diagram of the fish.

01:11 So we have the gravitational force acting downward and the tension force due to the wire acting upward on the fish.

01:20 Since the whole system is in equilibrium here, we can say that the tension force is equals to m .g.

01:28 Now let's use our equation 11 .10 from the young's book.

01:34 To find the stretch in wire when the fish is hanging from here.

01:42 We get the equation which says the young's modulus is equal to tensile stress over tensile strain.

01:51 The tensile stress is given by the tensile force over cross -sectional area and the tensile strain is given by the change in length over length.

02:01 So i can use this equation to find my change in length or the stretcher length.

02:06 Chin wire which gives me this equation m g l over a y so from here onwards i will use notations where i'll say that delta l as d l and i'll denote l not as l.

02:26 So we have all the unknowns here except for the youngs modulus so youngs modulus is basically a constant and for steel our wire is made of steel.

02:37 So for steel the young's model is 20 times 10 to the power 10 newtons for meter square.

02:43 So plugging all those values in this equation we get our stretch in wire to be 6 .62 times 10 to the power minus 4 meter.

03:08 I'll say this dl as dle e because this is the unstretched this is the stretching in the wire when the wire is at equilibrium so let's say this d le now let's move on to the next part so in the next part we have a slowly varying force f which is applied at the lower end of the wire and the change in length is given as delta l so my change in length because of the varying force is given to be 5 times 10 to the power minus 4 meter.

03:53 So in part b, it asks us what is the warp done by the gravitational force? so walk done is basically given by integral over f .d .x since my gravitational force is a constant, so this equation then gives f .d where d is the displacement and here this is the stretch in the wire.

04:19 I'm sorry.

04:22 So we have f as mass times gravity and dl as this value.

04:27 So plugging all these values into this equation we get my work done as 2 .2 times 10 to the power minus 2 joules.

04:45 So this is the walk done due to gravity when we have a varying force applied.

04:51 On the wire at the bottom.

04:55 So for the third question, the third question asks us to find the work done by the varying force that is being applied.

05:04 So for to start with, i'll again use equation 11 .10, which gives me my equation for young's modulus as a function of stress and strain.

05:18 So using this equation, i can find my force to be y times a times dl over l.

05:29 Now since the wire is getting stretched over here, i can assume that my wire, i can assume my wire to be a spring, to be acting like a spring.

05:40 And by hooke's law, we can say that the force or the spring force is equal to the spring constant k times the change in length...

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