# Physics question

## Recommended Posts

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 17 kg and the larger bottom crate has a mass of m2 = 92 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).

1)

The rope is pulled with a tension T = 248 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? 2.275 m/s^2

2) In the previous situation, what is the frictional force the lower crate exerts on the upper crate? 38.675 N

3) What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide? 854.56N

4) The tension is increased in the rope to 1223 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate? 6.076 m/s^2

5)

As the upper crate slides, what is the acceleration of the lower crate? this is the only one that I could not figure out

I know all my answers for 1-4 are correct but i cant seem to get the answer to 5. If someone could help me with an equation it'd be appreciated

edit:

QUESTION HAS BEEN RESOLVED

Edited by coach94
##### Share on other sites

I agree with the answers in the first parts.

Whilst the two boxes are moving together they act as a single object as far as the rope is concerned and the friction between them counts as an internal force, so your calculatuion does not show any frictional retarding force on the lower box.

As soon as they start to slide past each other the upper box exerts a retarding force on the lower box.

thus you need to calculate this and do a horizontal force balance to find the actual accelerating force on the lower box.

Does this help?

##### Share on other sites

• 2 years later...

Isn't the answer to #4 : 5.144 m/s^2 ?

If not, then how did you arrive at : 6.076 m/s^2 ?

Thanks

V V

##### Share on other sites

It is a question of patiently putting in the correct masses and friction coefficients at the appropriate points.

I don't get exactly the same results, but it would depend upon the value adopted for g.

Here are my calculations

1)
$acceleration = \frac{{Tension}}{{totalmass}} = \frac{{248}}{{92 + 17}} = 2.275m/{s^2}$
2)

$Force = mass*acceleration = 17x2.275N$

3)

$Max{F_{top}} = mass*\max acceleration$

$0.8*17*g = 17*{a_{\max }}$

${a_{\max }} = 0.8g = 0.8*9.81 = 7.848m/{s^2}$

$Max{F_{total}} = Mas{s_{total}}*{a_{\max }} = 109 *7.848 = 855.432N$

4)

${F_{topsliding}} = Mas{s_{top}}*acceleratio{n_{top}}$

$0.62*17*g = 17*{a_{topsliding}}$

${a_{topsliding}} = 0.62g = 0.62*9.81 = 6.0822$

5)

$acceleration = \frac{{netforce}}{{mas{s_{botom}}}} = \frac{{Tensionpull - frictionfromtop}}{{92}} = \frac{{1223 - 0.62*17*9.81}}{{92}}12.17m/{s^2}$

Edited by studiot

## Create an account

Register a new account