ps03sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Problem Set 3: Newton’s Laws Solutions Problem 1: Floor Mop The handle of a floor mop of mass m makes an angle ! with the vertical. Let μ be the coefficient of friction between the mop and the floor. Neglect the mass of the handle. a) Find the magnitude of the force F directed along the handle required to slide the mop with uniform velocity across the floor. b) If is smaller than a certain angle 0 , the mop cannot be made to slide no matter how great a force is applied. Find 0 .
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Problem 2: A particle of mass m enters a region horizontally at time t = 0 with speed v 0 midway between two plates that are separated by a distance h as shown in the figure. The particle is acting on by both gravity and by a time varying force that points upward and has magnitude ! F = bt where b is a positive constant that is sufficiently large such that the particle hits the top plate without ever touching the bottom plate. a) Sketch the motion of the particle. b) How long does it take for the particle to reach its lowest point? c) What is the minimum possible value of b such that the particle does not hit the lower plate? Solution: Initially only the gravitational force acts on the object but as the particle starts to move, the vertically upward force grows linearly in time. This means that the particle will start to move both forward and downward but will eventually slow reach a lowest point and then start to move forward and upwards until it hits the plate. A sketch of the motion is shown in Figure 4.1 Figure 4.1
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The lowest point in the orbit is reached when the vertical-component of the velocity is zero. The free body diagram for the forces acting on the object are shown in Figure 4.2a. We choose a coordinate system so the origin is located at the point of entry midway between the plates and positive y-axis points upwards (Figure 4.2b). Figure 4.2a Figure 4.2b Since the upward force grows linearly in time, the y-component of the acceleration must also increase linearly in time. We can integrate the y-component of the acceleration to find the y-component of the velocity and then integrate again to find the y-component of the position. This provides a model for the non-uniform motion. From our two equations we can determine the time interval, t b , it takes to reach the lowest point of the path when y-component of the velocity is zero, v y ( t b ) = 0 , and the minimum value of b such that the lowest point of the orbit satisfies y ( t b ) = ! h / 2 . The component equations of motion corresponding to Newton’s Second Law, ! F = m ! a , are in the y-direction: ˆ j : bt ! mg = ma y . (0.1) Thus the vertical acceleration is a y = b m t ! g (0.2) and the x-direction: ˆ i : 0 = ma x . (0.3) Thus the vertical acceleration is a y = b m t ! g (0.4) and the horizontal acceleration is zero: a x = 0 . (0.5) We can now integrate the y-component of the acceleration to get the y-component of the velocity
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v y ( t ) ! v y 0 = a y dt 0 t " = b m t ! g # $ % & ' ( dt 0 t " = b 2 m t 2 ! gt . (0.6) Since the initial y-component of the velocity is zero, Eq. (0.6) becomes v y ( t ) = b 2 m t 2 ! gt . (0.7)
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This note was uploaded on 04/13/2011 for the course PHYSICS 8.01 taught by Professor Guth during the Fall '09 term at MIT.

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ps03sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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