course Phy 201 ???????????`?y?assignment #025
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12:23:18 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 --> 35kg*9.8m/s^2=343N .30 * 343N=102.9N 343N -102.9N=240.1N Fnet=ma a=0
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12:25:10 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 --> The normal force=35lb weight of the crate I used the 35 as kg instead of lbs frictional force=.30*35lb=10.5lb
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12:50:01 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 --> Fgx=mg sin theta Fgy=-mg cos theta Fx=max mgsine theta-ukFn=max Fy=may Fn-mgcos theta=may=0 Fn=mgcos theta mg sine theta -uk(mgcos theta)=max g *sine 37deg -uk(g *cos 37degrees)=ax ax=g(.602)-ukg *.799 ax=9.8m/s^2 *.602 -uk *9.8m/s^2 *.799 ax=5.899 - uk*7.83 2.70m/s^2=5.899m/s^2 -uk*7.83m/s^2 Subtract 5.899m/s^2 -3.199m/s^2=-uk(7.83m/s^2) Divide by 7.83m/s^2 -.41=-uk .41=uk Frictional force= .41 *9.8m/s^2 *18kg*.799=57.79N
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13:00:35 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 --> I tried to follow the book on example 4.21 pg. 94 but didn't get the same answer. Force associated with the weight component moving the box downward=Fp= sine 37degrees *18kg*9.8m/s^2=106N Normal force is acting perpendicular to the gravitational force Fn=cos 37degrees *18kg*9.8m/s^2=141N Frictional force= Fnet-frictional coefficient*normal force=ma 106N-141N*Frictional coeff.=18kg*.270m/s^2 Frictional coefficient=.717
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13:00:40 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 -->
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13:00:43 ** 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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