course Phy 231 I posted another way explaining the fish problem at the end of the form in response the question posed by another student, and I know that you already worked it out earlier by re-reading the answer I posted below. I hope this can help some students with any misunderstandings for the tension problems. šiªÝdÃé÷¢èàHÎ郪›}¬Âœgé¦ïÈø¸«šèàassignment #013
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14:18:16 prin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider
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RESPONSE --> Ok confidence assessment: 3
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14:18:25 A force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m. Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg.
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RESPONSE --> Ok self critique assessment:
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14:18:38 prin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel
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RESPONSE --> Ok confidence assessment: 3
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14:18:52 ** The initial velocity of the bullet is zero and the final velocity is 175 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.11 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. **
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RESPONSE --> Ok self critique assessment:
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14:18:56 gen phy 4.08. breaking strength 22 N, accel 2.5 m/s^2 breaks line. What can we say about the mass of the fish?
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RESPONSE --> Ok confidence assessment: 3
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14:19:02 The fish is being pulled upward by the tension, downward by gravity. The net force on the fish is therefore equal to the tension in the line, minus the force exerted by gravity. In symbols, Fnet = T - m g, where m is the mass of the fish. To accelerate a fish of mass m upward at 2.5 m/s^2 the net force must be Fnet = m a = m * 2.5 m/s^2. Combined with the preceding we have the condition m * 2.5 m/s^2 = T - m g so that to provide this force we require T = m * 2.5 m/s^2 + m g = m * 2.5 m/s^2 + m * 9.8 m/s^2 = m * 12.3 m/s^2. We know that the line breaks, so the tension must exceed the 22 N breaking strength of the line. So T > 22 N. Thus m * 12.3 m/s^2 > 22 N. Solving this inequality for m we get m > 22 N / (12.3 m/s^2) = 22 kg m/s^2 / (12.3 m/s^2) = 1.8 kg. The fish has a mass exceeding 1.8 kg.
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RESPONSE --> Ok self critique assessment: 3
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14:34:21 univ phy 4.38*parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist?
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RESPONSE --> The weight of the parachutists is simply the force applied due to gravitational acceleration and the mass of the parachutists; therefore, the weight is 55 kg * 9.8 m/s^2 = 539 N if we treat down as the positive direction. The parachutists will then have an acceleration found by the difference in forces acting upon him/her, which is 539 N - 620 N = - 81 N, which indicates this much force in the upwards direction; therefore, acceleration is - 81 N/ 55 kg = - 1.47 m/s^2. confidence assessment: 3
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14:36:39 Describe the free body diagram you drew.
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RESPONSE --> My free body diagram shows a parachutists with a parachute deployed above them. It depicts a combined mass of 55 kg for both the parachute and the parachutist. It shows the acceleration of gravity acting downwards with a magnitude of 9.8 m/s^2, and the force of air resistance with magnitude 620 N in the upwards direction. It is from this set up I began deriving my earlier work. confidence assessment: 3
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14:37:05 The 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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RESPONSE --> Ok self critique assessment: 3
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14:47:43 univ phy (4.34 10th edition) fish on balance, reading 60 N when accel is 2.45 m/s^2 up; find true weight, situation when balance reads 30 N, balance reading when cable breaks What is the net force on the fish when the balance reads 60 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read when the cable breaks?
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RESPONSE --> I do not have the 10th edition; therefore, even though I was able to obtain some information such as the object having 24.5 kg for mass. I could not fully answer the problem given without the textbook's description of the event. confidence assessment: 3
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14:49:02 ** 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 Fnet is still m * a and we have 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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RESPONSE --> As I thought there was more information to the problem that given; therefore, even my mass value was not correct. self critique assessment: 3
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14:58:41 STUDENT 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.
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RESPONSE --> What this student did was is find the difference in the acceleration caused by gravity and the opposing ( in direction) acceleration of the fisherman. The actual application should be that tension on the rod would need to added to the acceleration of gravity, because the line of the pole feels two forces acting upon it: the force of gravity and the weight of the fish, and the force of the line being jerked upwards by the fisherman. The line will experience tension equal to the sum of all forces that are ""pulling"" on it. I don't now how this may sound from a mathematical perspective, but by using this logic and then setting up the problem we should find that the mass of the fish would be at least 22 N = ( 9.8 m/s^2 + 4.5 m/s^2) m(fish); therefore, 1.54 kg = m(fish). At least this is how I see the problem could have been worked differently. confidence assessment: 3
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15:01:07 ** 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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RESPONSE --> When I put in the value of 22/ 14.3 into my calculator I find not 1.8, but the 1.54 that I reported. Is this an mistyped value by you or do I need to be checking the validity of my calculator? self critique assessment: 3
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