Query Questions 26

#$&*

course Phy 241

November 15 around 4:30pm.

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I DECLARE DOWN TO BE POSITIVE!

Can't miss that declaration. Good.

Fnorm=4(9.807)(cos 30)=33.98N*(0.25)=8.495N

Fnorm=8(9.807)(cos 30)=67.94N*(0.35)=23.78N

Gravitational:

4(9.807)(sin 30)=19.614N

8(9.807)(sin 30)=39.228N

Fnet=ma and a=Fnet/m

4kg: 19.614-8.495=11.12N downward

8kg: 39.228-23.78=15.45N downward

Accelerations:

11.12/4kg=2.78m/s^2

15.45/8kg=1.931m/s^2

Total mass:

11.12+15.45=26.57N

Acceleration: 26.57/(8+4)=2.214m/s^2

I know tension will be:

T=Fnet-mgsin(theta)+mu*N

Would I use the total mass or just the 8kg???

That will depend on whether the two masses accelerate at the same rate, which might or might not be so if the system is reversed (see the given solution).

If the entire system accelerates together, then the net force acts on the total mass.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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

I saw how to solve for tension, using the 8kg*(total acceleration)=17.712N for Fnet. Got it!

------------------------------------------------

self-critique rating #$&*:

Good work.

#$&*