Query 13

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course PHY 231

10/16 5:44 pm

013.  `query 13*********************************************

Question:    `quniv phy  4.42 (11th edition 4.38) parachutist 55 kg with parachute, upward 620 N force.  What are the weight and acceleration of parachutist?

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Your solution: 

W=55*9.81 = 540 N

Fnet = 620-540 = 80 N

a = Fnet/m= 80/55 = 1.45 m/s^2

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Question:    `qDescribe the free body diagram you drew.

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Your solution: 

 The FBD is just the force up from the parachute and the force weight down and the Fup should be just slightly larger than the weight vector.

 

confidence rating #$&*:

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Given Solution: 

`aThe weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx..  So the parachutist experiences a downward force of 540 N and an upward force of 620 N.  Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is

-540  + 620 N = 80 N.

 Your free body diagram should clearly show these two forces, one acting upward and the other downward.  The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx..

 

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Self-critique (if necessary): OK

 Self-critique Rating: OK

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Question:    `quniv phy  (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator.  The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction.

What is the net force on the fish when the balance reads 50 N?

What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks?

 

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Your solution: 

 Found this problem in my 9th edition but the force was 60N in it. The way I see this is that the force of the elevator up and the weight pulling down add so that the 50N is the total force. 50 = 9.81*m + 2.45*m ---> m= 4.1kg The next part I use the same setup: 30N= 9.81(4.1)+4.1(a) -----> a = - 2.5 m/s^2 so the elevator is accelerating down the shaft and this makes sense the 9.81*4.1= 40.2N so the elevator has to 'lessen' the acceleration of gravity since, in this instance, the net acceleration is 9.81-2.45 = 7.4 m/s^2. If the cable breaks then the weight of the fish is no longer a factor as the force in the cable holding the fish to the balance is a reactive force so the scale reads 0.

  

confidence rating #$&*:9

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Given Solution: 

`a** Weight is force exerted by gravity.

Net force is Fnet = m * a.  The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity.

So m a = 50 N - m g, which we solve for m to get

m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg.

If the balance reads 30 N then

F_net = m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so

a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2.

If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2.  Net force will be -m g; net force is also Fbalance - m g.  So

-m g = Fbalance - m g and we conclude that the balance exerts no force.  So it reads 0. **

  

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Self-critique (if necessary): Well I looked at this differently but got the same results. Not sure if I need to change my thinking since my strategy may have only worked here and not in general.

The way to think of this one:

F_net = m a

F_net = 50 N - m g

Set the two expressions for F_net equal to one another, and solve the resulting equation for m.

m a = 50 N - m g

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self-critique rating #$&*: OK

  `qSTUDENT QUESTION:

   I had trouble with the problems involving tension in lines. For example the Fish prob.

Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish.

Here's what I did.

Sum of F = Fup + F down

-22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish)

-22N = -5.3 m/s^2 m(fish)

m(fish) = 4.2 kg

I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17.

 Given Solution: 

`a** Think in terms of net force.

The net force on the fish must be Fnet = m a = m * 4.5 m/s^2.

Net force is tension + weight = T - m g, assuming the upward direction is positive.  So

T - m g = m a and

T = m a + m g.  Factoring out m we have

T = m ( a + g ) so that

m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx..

The same principles apply with the elevator.  **

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Self-critique (if necessary): I don't know what this is referring to about the fishing lines. I can't find it in the book or problem sets.

This is an old problem, from a previous edition. However it is pretty much identical to the first-and-elevator problem.

A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks. What is the mass of the fish?

in this case setting expressions for net force equal we get

T - m g = m a

In this case we know that T >= 22 N, so

m a + m g >= 22 N and

m >= 22 N / (a + g).

 Self-critique rating:

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&#Good responses. See my notes and let me know if you have questions. &#

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