Idea Transcript
Discussion Section Review Problems J.K. Brosch Department of Physics and Astronomy, University of Delaware Sharp Lab, 102 January 14, 2016
Chapter 5: Question 3 If the 1 kg standard body has an acceleration of 2 m/s2 at 20.0◦ to the positive x-axis, what are (a) the x component of the net force acting on the body, (b) the y component of the net force acting on the body, (c) and what is the net force in unit vector notation? Solution: To solve this problem, we only need to apply Newton’s Second Law, F = ma, in each of the desired direction (a) In the x directions, the x component is given by the cosine of the angle, i.e. Fx = max = ma cos(20◦ ) = 2 cos(20◦ ).
(1)
(b) In the y directions, the y component is given by the sine of the angle, i.e. Fy = may = ma sin(20◦ ) = 2 sin(20◦ ).
(2)
(c) In unit vector notation, it will simply be the results in (1) - (2) with their respective unit vectors, i.e. F = ma = Fx xˆ + Fy yˆ = 2 [cos(20◦ )ˆ x + sin(20◦ )ˆ y] .
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Chapter 5: Question 6 In a two dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on a tire at the angles shown in the image below. The tire remains stationary in spite of the three pulls. Alex
pulls with a magnitude |FA | = 220 N , and Charles pulls with a magnitude of |FC | = 170 N . Note that the direction of FC is not given. What is the magnitude of Betty’s FB ? Solution: Since the tire is stationary, we must have by Newton’s Second Law that F = FA + FB + FC = ma = 0, where of course here the 0 is understood to represent the zero vector, 0 = 0ˆ x +0ˆ y. Examining the free-body diagram below,
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we have, 0=
X
Fx = Fc cos φ − FA cos θ
(3)
0=
X
Fy = FA sin θ + FC sin φ − FB .
(4)
To solve for FB in (4), we first need to find φ. Using our values for FC and FB together with the fact that θ = 47◦ (why? check this for yourself), we have cos φ =
FA cos θ = 0.883, FC
and thus φ = 28.◦ . Substituting φ into (4), we have FB = FA sin θ + FC sin φ = 241 N.
Chapter 5: Question 11 A 2 kg particle moves along the x-axis, being propelled by a variable force along the axis. It’s position is given by x = 3 + 4t + ct2 − 2t3 , (5) where c is a constant. At t = 3, the force on the particle has a magnitude of 36 N , and is in ˆ direction. What is the constant c? the negative x Solution: For this problem, we need to find the acceleration; to do this we differentiate (5) twice to get d v(t) = x(t) = 4 + 2ct − 6t2 , dt and d d2 a(t) = v(t) = 2 x(t) = 2c − 12t. (6) dt dt We now substitute (6) into Newton’s Second Law to get F = ma = 2 (2c − 12t) . Since at time t = 3, F = −36 N , we have −36 = 4c − 72. We now solve this for c and obtain c = +9 m/s2 .
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Chapter 5: Question 14 A block with a weight of 3 N is at rest on a horizontal surface. A 1 N upward force is applied to the block through a vertical string. Draw the free-body diagram. What are the (a) magnitude (b) and direction of the force of the block on the horizontal surface? Solution: There are a total of three vertical forces acting on the block; draw the free-body diagram to illustrate this (left as an exercise to the student). The three forces are: (i) the force of gravity pulling downwards on the block, with force 3 N , (ii) the string pulling up on the block with force 1 N , (iii) and the normal force FN . Since there is no acceleration, we have X Fy = FN + 1 N − 3 N = 0. Solving this for FN now gives us FN = 2 N . (a) Using Newton’s Third Law, the force exerted by the block on the surface has the same magnitude but the opposite direction, i.e. FB→S = 2 N. (b) Since it has opposite direction, the force is in the downwards direction.
Chapter 5: Question 16 Some insects can walk below a thin rod by hanging from it. Suppose that such an insect has mass m and hangs from the rod at an angle θ, as seen in the figure below.
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Its six legs are all under the same tension, and the leg sections nearest the body are all horizontal. (a) What is the ratio of the tension in each tibia (forepart of the leg) to the insects weight? (b) If the insect straightens out its legs somewhat, does the tension in each tibia increase, decrease, or stay the same? Justify your answer. Solution: (a) There are six legs, and the vertical component of the tension force in each leg is T sin θ. For vertical equilibrium, Newton’s Second Law tells us that 6T sin θ = mg, which thus gives us T =
mg . 6 sin θ
(b) The angle θ is measured from the horizontal, so as the insect straightens out its legs, θ will increase, i.e. get closer to π2 . In doing so, this causes sin θ to increase as well. Why does this happen? Be sure that you can give a justification of this. Thus, T will decrease. Again, why is this true? Be sure to justify your answer using the result from part (a).
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