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Fall 07 Midterm 2 Solutions Ch. 4-5, 10.5-6

Fall 07 Midterm 2 Solutions Ch. 4-5, 10.5-6 - 1 Forces(8...

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Unformatted text preview: 1. Forces (8 pts) a. If the action and reaction forces described by Newton's Third Law are always equal in magnitude and opposite in direction, doesn't the net vector force on any object always add up to zero? Explain your answer. (4 pts) The action and reaction forces described by Newton's Third Law are equal in magnitude and opposite in direction, but they are always exerted on different objects. Thus, if object A exerts a (vector) force F on object B, then object B exerts a (vector) force F on object A. But this does not mean that the net vector force on A or B is zero. b. Astronauts orbiting on the Space Shuttle (about 90 miles above the Earth) are "weightless" they can float through the cabin, and objects released in midair do not fall to floor. Explain this phenomenon using Newton's Laws of Motion and/or Gravitation. (4 pts) The gravitational force on objects (including astronauts) in low Earth orbit is not much smaller than on the ground (since 90 miles is a practically negligible distance compared to the radius of the Earth). They are "weightless" because they and the other objects in their spacecraft have a large speed, and travel around the planet in (approximately) uniform circular motion. Thus, when an object is released in the cabin, its inertia causes it to continue in its (approximately) circular orbital path, with the gravitational force providing the necessary centripetal acceleration. A 2. Dynamics (8 pts) a. An object of starts at rest, with position x = 0 at time t = 0. It is subjected to a constant force F = 7.00 N, and at t = 5.00 s, its position is x = +9.00 m. What is the object's mass? Under a constant force, the object accelerates uniformly from rest, and its position is described by: x = a t2 = (F/m) t2 This can be rearranged to give: m = F t2/ (2x) = (7.00 N) (5.00 s)2 / (2 (9.00 m)) = 9.72 kg 9.72 kg b. Two blocks with mass 3.00 kg are connected by a massless string. One block is pulled up a frictionless plane inclined at 35.0 from the horizontal by an external force, with the second block pulled behind it (by the string). If the tension in the string connecting the blocks is 80.0 N, what is the magnitude of the external force? Work in a coordinate system where x is up the slope and y is normal to the slope. For the second block, the net force in the x direction is T Fx = T mg sin So the acceleration of this block (and the other) is mg ax = Fx/m = T/m g sin mg sin For the first block, the force in the x direction is F F'x = F T mg sin = m ax = m(T/m g sin), where F is the external force we are trying to find. T Thus F = 2T = 160. N mg mg sin 160. N A 3. Friction and Velocitydependent Forces (8 pts) a. An object slides along a horizontal surface with a coefficient of kinetic friction of 0.350. If the object's initial velocity is 2.00 m/s, how far does it travel before stopping? (4 pts) The frictional force on the object is given by: f = k |n|, where n is the normal force between the object and the surface. Since the surface is horizontal and the object does not accelerate vertical, n mg = Fy = 0 n = mg f = k mg The (negative) acceleration due to friction is therefore a = f/m = k g The object's final velocity (when stopped) is 0, and its initial velocity v0 is given. vf2 v02 = v02 = 2 a x Substituting, x = v02/(2kg) = v02/(2kg) = (2.00 m/s)2/[2 (0.350) (9.80 m/s2)] = 0.583 m 0.583 m b. An object which starts from rest is subject to a velocitydependent force F = b v as it falls due to gravity. Explain what happens and why (i.e. describe the motion in words, 3 pts) and sketch a graph (1 pt) of the object's speed as a function of time. (4 pts) Since the object starts from rest, the resistive force is initially zero, and the object begins free fall. As its velocity increases, the resistive force increases, and its acceleration is less than g. Eventually, the object gains sufficient velocity so that the resistive force is equal in magnitude to its weight. When this occurs, the net force on the object is zero and it maintains a constant ("terminal") velocity afterward. v t A 4. Circular Motion A ball is tethered to the top of a vertical pole by a massless cord of length L = 3.00 m. When the ball is at rest, it barely touches the ground (i.e. the pole is also 3.00 m long). When the ball is struck or thrown, it travels in a circle some distance h above the ground. (8 pts) a. If the ball makes 0.750 revolutions around the pole per second, what is its height h above the ground? (4 pts) If the ball is height h above the ground, (Lh)/L = cos and L cos = Lh where is the angle of the cord from the vertical. L The radius of the circle the ball travels in is then R = L sin The forces on the ball are its weight and the tension in the cord: R Fx = T sin Fy = T cos mg If the ball maintains a constant height, Fy = 0, so T cos = mg If the ball makes f revolutions per second, its speed is v = f (2R) T Fx must supply the required centripetal acceleration: Fx = T sin = mv2/R = m (42 f2 R2) /R = 42m f2 R = 42mf2L sin Substitute for T: (mg/cos) sin = 42mf2 L sin mg Cancel the factor m sin on both sides and isolate L cos = L h L cos = L h = g/(42f2), h = L g/(42f2) 2.56 m h = (3.00 m) (9.80 m/s2)/[4 (3.1416)2(0.750/s)2) = 2.56 m (Essentially the same as homework Problem 5.19) b. What is the tension in the cord? (4 pts) The mass of the ball was not given (an oversight), so this part was graded leniently. From above, T = mg/cos = mg/[(Lh)/L] = mgL/(Lh) = (66.6 N/kg) m (kg) It is also possible to use the equation for Fx: T sin = m v2/R = m (4 2 f2 R2)/R = 42mf2R = 42mf2L sin, so T = 42mf2L = (66.6 N/kg) m (kg) Thus, if the mass of the ball were 1.00 kg, the tension would be 66.6 N. Lh h A 5. Torque and Vector Product a. A force F = (5.00 N j 2.00 N k) is applied to an object at a position r = (3.00 m i + 4.00 m j) from the object's axis of rotation. Find the resulting torque vector. = r F x = ryFz rzFy = (4.00 m)(2.00 N) 0 = 8.00 N m y = rzFx rxFz = 0 (3.00 m)(2.00 N) = +6.00 N m z = rxFy ryFx = (3.00 m)(5.00 N) 0 = 15.0 N m (8.00 i + 6.00 j + 15.0 k) N m b. When a block of wood on a table is pushed near the top, it will tip over, but when pushed close to the bottom, the block will slide (if the force from the pencil exceeds the maximum force of static friction). Consider the "just right" situation in between, where the pencil is at height h and the block is in equilibrium, between tipping and sliding. If the pencil is moved up from h, the block will tip over, and if the pencil is moved down from h, it will slide. At the "just right" position where the block is on the verge of tipping, all forces on the block from the table are exerted at the axis of rotation, as shown. Find the coefficient of static friction in terms of h and the block's thickness w. Call the external force of the pencil F. The forces on the block are: Fx = F f = F sn = 0 (equilibrium) Fy = n mg = 0 (equilibrium) So n = mg, and F = smg Relative to the axis at the corner of the block, the torque is: = (w/2 i + H/2 j)(mg j)+(w i + h j)(F i) = 0 (equilibrium) The block's weight acts at the center of the block, r = w/2 i + H/2 j from the axis of rotation. The height H of the block is unknown, but doesn't affect the answer. (j j = 0) Friction and the normal force from the table act at the axis of rotation, and create no torque z = wmg/2 hF = 0 Substitute F = smg: wmg/2 = h (smg), so s = w/(2h) w/(2h) A ...
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