Query 26

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course PHY 231

11/09 12:42 pm

026. `query 26*********************************************

Question: `qUniv. 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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Your solution:

Wa_||= 4(9.81)(sin 30)= 19.6 N

Wa_perp= 4(9.81)(cos 30) = -33.98---->Fna= 33.98---->ƒa=µa*Fna= -0.25(33.98)= -8.5 N

Wb_||= 8(9.81)(sin 30)= 39.24 N

Wb_perp= 8(9.81)(cos 30)= -67.97 N----->Fnb= 67.97---->ƒb= µb*Fnb= -0.35(23.79)= -23.79 N

Fnet_y= Wa_||-ƒa+Wb_||-ƒb= 19.6-8.5+39.24-23.79 = 26.55 N

a) a_sys= 26.55/12 = 2.21 m/s^2

b) Fnet_a = 4(2.21) = Wa_||-ƒa -T= 19.6 - 8.5 -T---> T= 2.2 N

Fnet_b= 8(2.21) = Wb_||-ƒb + T= 39.24-23.79 +T ------> T=2.2 N

c) If 4 kg block is above the 8 kg block then T = 0 and eventually 4 kg will hit 8 kg. Their a's are 1.93 and 2.78 m/s^2 for b and a respectively. With a little more info you could find far/long it would be before they collide and and what happens after they collide.

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When the two blocks collide and move as one, what would µ then be? Does it add or do you assume the greater one, or the one in front?

the blocks do not combine into a single solid block, so each still experiences the force calculated based on its normal force and coefficient of friction.

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Confidence rating:

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Given Solution:

`a** 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. **

Self-critique (if necessary): OK

Self-critique Rating:

&#This looks good. Let me know if you have any questions. &#

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