asst 26 query

course Phy 121

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Physics I

07-15-2007

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14:42:17

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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I tried to follow the example in the book for section 4.8. I did 35kg/9.8 = 3.57N. Then I multiplied this by the frictional force of .30 to get 1.07N. I think that this is the force that has to be overcome to push the box against the frictional force.

The units of your calculation would be kg / (m/s^2) = kg * s^2 / m. This is not consistent with the unit Newtons, which is kg * m/s^2.

Your error was to divide the mass by the acceleration of gravity, which doesn't give a meaningful result. You should have multiplied.

However except for this error your overall idea was correct.

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14:43:31

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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Why is the frictional force in pounds here and not it Newtons and why doesn't gravity come into account at all here?

The original problem in the text had a 35 lb crate. Since lb is a measure of force, not mass, the normal force would then be 35 lb and the frictional force is 0.30 * 35 lb = 10.5 lb.

However at some point the book changed the force to kg and I apparently didn't notice the change in units, so the solution here is still given in terms of a 35 lb weight.

If the mass of the box is 35 kg, then since gravity would accelerate this mass at 9.8 m/s^2 if it was free to fall, gravity must be exerting a force of 35 kg * 9.8 m/s^2 = 340 N, approx.. That is, the weight of the box is about 340 N.

If the box is resting on or sliding across the level floor, the floor must exert a normal force of 340 N in order to counter the force of gravity and prevent the box from accelerating downward.

In this case the frictional force is 0.30 times the normal force, or 0.30 * 340 N = 102 N, approx.

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14:43:34

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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14:43:36

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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14:43:38

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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14:43:39

** 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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See my note on the text problem, which differed from the problem solved in the given solution, and let me know if you have questions.