course phy201 ????????F??????assignment #026?S??t???}??`?R??Physics I
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21:57:42 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 --> Ffr = .3 * 15.9Kg*g = 47N
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21:58:48 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 --> got the same answer but in N
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22:22:37 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 --> a=Fnet/m .27m/s^2 =Fnet/18Kg Fnet = .27m/s^2*18Kg = 4.9N Ffr=4.9N-106N = -101N Fn=18Kg*g*sin37deg = 106N coefficient of friction = 101/106 =.95
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22:25:10 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 --> should have used cos for Fn not sin. But I understand the solution.
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