course Phy 201 ????????~M}P??Student Name: assignment #023
.................................................
......!!!!!!!!...................................
20:34:48 `q001. Note that this assignment contains 3 questions. . A chain 200 cm long has a density of 15 g/cm. Part of the chain lies on a tabletop, with which it has a coefficient of friction equal to .10. The other part of the chain hangs over the edge of the tabletop. If 50 cm of chain hang over the edge of the tabletop, what will be the acceleration of the chain?
......!!!!!!!!...................................
RESPONSE --> The length of chain hanging over the edge of the table is 50cm * .015kg/cm=.75kg. This is being acted upon by the acceleration of gravity 9.8m/s^2: .75kg*9.8m/s^2=7.35N. The frictional force on the chain is due to the 150cm still on the table. This is acted upon by gravity as well: 2.25kg*9.8m/s^2=22.05N. The friction is .10 of the force: 2.21N due to friction. This is subtracted from the force of acceleration on the suspended 50 cm of chain: 7.35N-2.21N=5.14N. Since a=F/m, a=5.14N/3kg=1.7m/s^2.
.................................................
......!!!!!!!!...................................
20:36:40 The part of the chain hanging over the edge of the table will experience an unbalanced force from gravity and will therefore tend to accelerate chain in the direction of the hanging portion. The remainder of the chain will also experience the gravitational force, but this force will be countered by the upward force exerted on the chain by the table. The force between the chain and the table will give rise to a frictional force which will resist motion toward the hanging portion of the chain. If 50 cm of chain hang over the edge of the tabletop, then we have 50 cm * (15 g/cm) = 750 grams = .75 kg of chain hanging over the edge. Gravity will exert a force of 9.8 meters/second ^ 2 * .75 kg = 7.3 Newtons of force on this mass, and this force will tend to accelerate the chain. The remaining 150 cm of chain lie on the tabletop. This portion of the chain has a mass which is easily shown to be 2.25 kg, so gravity exerts a force of approximately 21 Newtons on this portion of the chain. The tabletop pushes backup with a 21 Newton force, and this force between tabletop and chain results in a frictional force of .10 * 21 Newtons = 2.1 Newtons. We thus have the 7.3 Newton gravitational force on the hanging portion of the chain, resisted by the 2.1 Newton force of friction to give is a net force of 5.2 Newtons. Since the chain has a total mass of 3 kg, this net force results in an acceleration of 5.2 N / (3 kg) = 1.7 meters/second ^ 2, approximately.
......!!!!!!!!...................................
RESPONSE --> ok. my numbers were slightly different, but our process was the same.
.................................................
......!!!!!!!!...................................
20:36:50 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
......!!!!!!!!...................................
RESPONSE --> to solve this, we substitute L for the length of chain hanging over the edge. The weight of this section is: L*.015kg/cm*9.8m/s^2=L*.147N/cm. The force acting on the other section of the chain (on the table) is: (200-Lcm)*.015kg/cm*9.8m/s^2=.147N/cm(200-Lcm). The frictional force on the chain is 1/10 this amount: .147N/cm(200-Lcm) *.10= .0147N/cm(200-Lcm) These two forces must equal each other if the chain is to not move: .0147N/cm(200-Lcm)=.147N/cm(Lcm). L=18.2cm
.................................................
......!!!!!!!!...................................
20:40:27 The maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain. If x stands for the length in cm of the portion of chain hanging over the edge of the table, then the mass of the length is x * .015 kg / cm and it experiences a gravitational force of (x * .015 kg / cm) * 9.8 m/s^2 = x * .147 N / cm. The portion of chain remaining on the tabletop is 200 cm - x. The mass of this portion is (200 cm - x) * .015 kg / cm and gravity exerts a force of (200 cm - x) * .015 kg / cm * 9.8 meters/second ^ 2 = .147 N / cm * (200 cm - x) on this portion. This will result in a frictional force of .10 * .147 N / cm * (200 cm - x) = .0147 N / cm * (200 cm - x). Since the maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain, we set the to forces equal and solve for x. Our equation is .0147 N / cm * (200 cm - x) = .147 N/cm * x. Dividing both sides by .0147 N/cm we obtain 200 cm - x = 10 * x. Adding x to both sides we obtain 200 cm = 11 x so that x = 200 cm / 11 = 18 cm, approx..
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:56:34 `q003. The air resistance encountered by a certain falling object of mass 5 kg is given in Newtons by the formula F = .125 v^2, where the force F is in Newtons when the velocity v is in meters/second. As the object falls its velocity increases, and keeps increasing as it approaches its terminal velocity at which the net force on the falling object is zero, which by Newton's Second Law results in zero acceleration and hence in constant velocity. What is the terminal velocity of this object?
......!!!!!!!!...................................
RESPONSE --> the force of gravity of the object is: 5kg*9.8m/s^2=49N. Since terminal velocity will occur when air resistance becomes negligent due to the force of velocity, F=.125v^2 must equal the force of gravity=49N. This means that v=19.8m/s.
.................................................
......!!!!!!!!...................................
20:57:46 Only two forces act on this object, the downward force exerted on it by gravity and the upward force exerted by air resistance. The downward force exerted by gravity remains constant at 5 kg * 9.8 meters/second ^ 2 = 49 Newtons. When this force is equal to the .125 v^2 Newton force of friction the object will be at terminal velocity. Setting .125 v^2 Newtons = 49 Newtons, we divide both sides by .125 Newtons to obtain v^2 = 49 Newtons/(.125 Newtons) = 392. Taking square roots we obtain v = `sqrt (392) = 19.8, which represents 19.8 meters/second.
......!!!!!!!!...................................
RESPONSE --> ok
................................................."