course Phy 121 ??h?P???????Y??|??Student Name:
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12:59:23 `q001. Note that this assignment contains 3 questions. . Water has a density of 1 g per cm^3. If an object is immersed in water, it experiences a buoyant force which is equal to the weight of the water it displaces. Suppose that an object of mass 400 grams and volume 300 cm^3 is suspended from a string of negligible mass and volume, and is submerged in water. If the mass is suspended in equilibrium, what will be the tension in the string?
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RESPONSE --> d = 1g/cm^3 vol = 300cm^3 mass of water = (1g/cm^3)(300cm^3) = 300g = 0.3kg F = m*g = 0.3kg * 9.8cm/s^2 = 2.94 N upward mass of object = 400 g = 0.4kg F = m*g = 0.4kg * 9.8cm/s^2 = 3.92 N downward The net force includes the force pulling the object downward, the buoyance pushing it upward, and also the tension on the string. Since the object is suspended in equilibrium, we know the total force will be zero. Fnet = 2.94N - 3.92N + T = 0 T = -0.98 Newtons
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13:00:20 The 400 g mass will experience a downward gravitational force of .4 kg * 9.8 meters/second^2 = 3.92 Newtons. It will also experience in upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. This volume of water, at 1 g per cm^3, will have a mass of 300 grams and therefore a weight of .3 kg * 9.8 meters/second^2 = 2.94 Newtons. The forces acting on the mass are therefore the downward 3.92 Newtons of gravity, the upward 2.94 Newtons of the buoyant force and the tension, which we will call T, in the string. Since the system is in equilibrium these forces must add up to 0. We thus have -3.92 Newtons + 2.94 Newtons + T = 0, which has solution T = .98 Newtons. In common sense terms, gravity pulls down with 3.92 Newtons of force and the buoyant force pushes of with 2.94 Newtons of force so to keep all forces balanced the string must pull up with a force equal to the .98 Newton difference.
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RESPONSE --> I understand.
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13:01:48 `q002. A solid cylinder has a cross-sectional area of 8 cm^2. If this cylinder is held with its axis vertical and is immersed in water to a depth of 12 cm, what will be the buoyant force on the cylinder?
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RESPONSE --> The volume of the object submerged is equal to the volume of water displaced. A = 8cm^2 h = 12cm Volume = A * h = (8cm^2)(12cm) = 96cm^3 Since we are talking about water, we know that 1cm^3 = 1 gram. Therefore, there are 96 grams or 0.096 kilograms of water removed. The buoyant force is equal to the gravitational force of the water. Buoyant force = m*g = (0.096kg)(9.8cm/s^2) = 0.941 Newtons.
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13:02:33 At a depth of 12 cm, the volume of the immersed portion will be 12 cm * 8 cm^2 = 96 cm^3. This portion will therefore displace 96 grams of water. The weight of this displace water will be .096 kg * 9.8 meters/second^2 = .94 Newtons. This will be the buoyant force on the cylinder.
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RESPONSE --> I understand.
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13:05:08 `q003. The solid cylinder in the preceding problem has a total length of 18 cm and a mass of 80 grams. If the cylinder is immersed as before to a depth of 12 cm then released, what will be the net force acting on it at the instant of release?
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RESPONSE --> L = 18cm m = 80 g = 0.08kg h = 12cm The force pulling the object down is m*g. (0.08kg)(9.8cm/s^2) = 0.784 Newtons. The buoyancy of the water is still 0.941 Newtons. There are no other significant forces, so Fnet = 0.941N - 0.784N = 0.157 Newtons.
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13:08:13 The buoyant force on the cylinder is still .94 Newtons, directed upward. Gravity exerts a downward force of .080 kg * 9.8 meters/second^2 = .79 Newtons, approximately. The net force on the cylinder is therefore .94 N - .79 N = .15 N, directed upward. This will give its 80 gram mass and acceleration a = F / m = .15 N / .080 kg = 1.875 m/s^2. Note that as the cylinder rises less and less of its volume is submerged, so the buoyant force will decrease while the weight remains the same. Until the buoyant force has decreased to become equal and opposite to the weight, the net force will continue to be upward and the cylinder will continue to gain velocity. After this instant the cylinder will continue to rise, but the net force will be downward so that the cylinder will begin slowing down. Eventually the cylinder will come to rest and the net downward force will cause it to start descending once more. It will continue descending until the net force is again 0, at which the time it will have a downward velocity that will carry it beyond this point until it again comes to rest and the cycle will start over again.
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RESPONSE --> I did not calculate the acceleration of the mass because it did not ask for it. However, I do know how to find this information and I have made note about the decreasing buoyancy as time passes.
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???????????????assignment #026 026. `query 26 Physics I 07-12-2006
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23:02:19 gen phy and principles of phy 4.36: If the coefficient of kinetic friction is .30, how much force is required to push a 35 lb crate across the floor at constant speed?
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RESPONSE --> The book says that the crate is 35 kilograms, not pounds. m = 35kg uk = 0.3 The force required to push the crate is equal to the frictional force between the floor and the crate. f = 35kg * 0.3 = 10.5 kg
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23:02:59 If the crate is moving then the force exerted by friction is .30 times the normal force between it and the floor. If the push is horizontal, then the only horizontal forces acting on the crate are the downward force of gravity and the upward force exerted by the floor. Since the crate is not accelerating in the vertical direction, these forces are equal and opposite so the normal force is equal to the 35 lb weight of the crate. The frictional force is therefore f = .30 * 35 lb = 10.5 lb.
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RESPONSE --> I understand this problem.
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23:06:10 gen phy 4.55 18 kg box down 37 deg incline from rest, accel .27 m/s^2. what is the friction force and the coefficient of friction?
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RESPONSE --> m = 18kg a = 0.27m/s^2 theta = 37 degrees If the incline is at 37 degrees, the weight vector will be 270 degrees more than that. So the angle of the weight vector is 37deg + 270deg = 307 degrees. The normal force is equal to the gravitational force exerted on the box. FNorm = 18kg*9.8cm/s^2 = 176.4 Newtons Therefore, the x and y weight components are: X comp = 176.4N * cos(307deg) = 106.2 N Y comp = 176.4N * sin(307deg) = -140.9 N I know that the coefficient of friction applies to the normal force. The normal force is equal and opposite to the parallel weight component.
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23:07:12 GOOD STUDENT SOLUTION: (I don't know why, but I was hoping you would pick an odd numbered problem here)Here goes.....For an 18kg box on an incline of 37 degrees with an acceleration of .270 m/s/s, I first drew out a diagram showing the forces involved. Next the forces had to be derived. First, to find the force associated with the weight component parrallel to the inline moving the box downward....Fp=sin 37 deg(18kg)(9.8m/s/s)=106N. Next, the Normal force that is counter acting the mg of the box is found by.. Fn=cos 37 deg. (18kg)(9.8 m/s/s) = 141N. The frictional force can be found by using F=(mass)(acceleration) where (Net Force)-(frictional coeffecient*Normal Force)=(m)(a) so that... 106N - (141N * Friction Coeff.) = (18kg)(.270 m/s/s) where by rearranging, the frictional coeffecient is seen to be .717. INSTRUCTOR COMMENT: Good solution. Note that you should specify an x axis oriented down the incline, so that the acceleration will be positive. The weight vector being vertical in the downward direction is therefore in the fourth quadrant, at an angle of 37 degrees with respect to the negative y axis. Thus the weight vector makes angle 270 deg + 37 deg = 307 deg with the positive x axis and its x and y components are wtx = 18 kg * 9.8 m/s^2 * cos(307 deg) = 106 N and wty = 18 kg * 9.8 m/s^2 * sin(307 deg) = -141 N. You get the same results using the sin and cos of the 37 deg angle. The only other y force is the normal force and since the mass does not accelerate in the y direction we have normal force + (-141 N) = 0, which tells us that the normal force is 141 N. This also agrees with your result. **
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RESPONSE --> Could you tell me how to find the frictional force and the coefficient of friction?
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23:08:06 Univ. 5.90 (5.86 10th edition). 4 kg and 8 kg blocks, 30 deg plane, coeff .25 and .35 resp. Connected by string. Accel of each, tension in string. What if reversed?
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RESPONSE --> I am not required to answer this question.
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23:08:24 ** We will use the direction down the incline as the positive direction in all the following: The normal forces on the two blocks are 4 kg * 9.8 m/s^2 * cos(30 deg) = 34 N, approx., and 8 kg * 9.8 m/s^2 * cos(30 deg) = 68 N, approx. If sliding the 4 kg block will therefore experience frictional resistance .25 * 34 N = 8.5 N, approx. and the 8 kg block a frictional resistance .35 * 68 N = 24 N, approx. The gravitational components down the incline are 4 kg * 9.8 m/s^2 * sin(30 deg) = 19.6 N and 8 kg * 9.8 m/s^2 * sin(30 deg) = 39.2 N. If the blocks were separate the 4 kg block would experience net force 19.6 N - 8.5 N = 11.1 N down the incline, and the 8 kg block a net force of 39.2 N - 24 N = 15.2 N down the incline. The accelerations would be 11.1 N / (4 kg) = 2.8 m/s^2, approx., and 15.2 N / (8 kg) = 1.9 m/s^2, approx. If the 4 kg block is higher on the incline than the 8 kg block then the 4 kg block will tend to accelerate faster than the 8 kg block and the string will be unable to resist this tendency, so the blocks will have the indicated accelerations (at least until they collide). If the 4 kg block is lower on the incline than the 8 kg block it will tend to accelerate away from the block but the string will restrain it, and the two blocks will move as a system with total mass 12 kg and net force 15.2 N + 11.1 N = 26.3 N down the incline. The acceleration of the system will therefore be 26.3 N / (12 kg) = 2.2 m/s^2, approx.. In this case the net force on the 8 kg block will be 8 kg * 2.2 m/s^2 = 17.6 N, approx.. This net force is the sum of the tension T, the gravitational component m g sin(theta) down the incline and the frictional resistance mu * N: Fnet = T + m g sin(theta) - mu * N so that T = Fnet - m g sin(theta) + mu * N = 17.6 N - 39.2 N + 24 N = 2.4 N approx., or about 2.4 N directed down the incline. The relationship for the 4 kg mass, noting that for this mass T 'pulls' back up the incline, is Fnet = m g sin(theta) - T - mu * N so that T = -Fnet + m g sin(theta) - mu * N = -8.8 N + 19.6 N - 8.5 N = -2.3 N. equal within the accuracy of the mental approximations used here to the result obtained by considering the 8 kg block and confirming that calculation. **
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