Please slove this and GIVE DETAILS!!?

Question

I know the answer(from back of book) but I can't solve for it please help

a slab of mass m1 = 21 kg rests on a frictionless floor, and a block of mass m2 = 9 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.20. The block is pulled by a horizontal force of magnitude 58 N. What is the acceleration of the slab?

Ok this is the answer, maybe this will help

-0.840857142857 m/s^2

ukmudgal

THANKS THAT HELPED

Answer 1

The limiting friction on block=coefficient of static friction x normal reaction =0.3 x9 x 9.8=26.46 N

As the applied force 58N is greater than limiting friction, the block slides and accelerates.The force of friction 'f' on the block is kinetic friction= coefficient of kinetic friction x normal reaction =0.2 x9 x 9.8=17.64 N

The force on the slab is the force of friction that the block exerts on the slab which is 17.64 N

Acceleration of the slab=force / mass of the slab=17.64/21=0.84 m/s^2
Acceleration of the slab is 0.84 m/s^2

Answer 2

Interesting question!

[EDIT: Note that even though ukmudgal came up with the right answer, his analysis was wrong. He said: "As the applied force 58N is greater than limiting friction, the block slides and accelerates." It is true that the block accelerates, but whether or not it slides against the slab depends also on the mass of the slab. If the slab were sufficiently light, the static friction would be enough to make the slab accelerate at the same rate as the block. As it turns out, the slab is too heavy for that to happen.]


Whenever you have a system with two (or more) pieces that may (or may not) be moving at different rates, I find it's best to look at one piece at a time and write equations for all the forces acting on THAT one piece.

Some of those equations will contain unknown quantities; but usually by the time you combine them all the unknowns disappear.

Also, it's a good idea, when forces are pulling in opposite directions, to decide which direction should be called "positive" and which "negative" (doesn't matter which you choose, but you have to make a choice and stick with it, otherwise your minus signs will get mixed up).

So, let's pretend the horizontal force pulling the small block is towards the right; and let's call that direction "positive".

Now for the forces:

Forces on m1 (the slab):
---------------------------------
1) friction from the block. This COULD be either static or dynamic friction (we don't know yet, since we don't know whether the block is _sliding_ against the slab or not.
2) No other forces! (there's no friction between slab & ground)

(Note: we don't yet know which coefficient, 0.20 or 0.30 to use for the friction! But don't worry about that yet.)

So, the NET FORCE on the heavy slab is:

F1_net = friction
(This is toward the right; so it's positive)

Forces on m2 (the block):
-----------------------------------
1) friction with the slab. This is the same amount as the friction on m1, but it pulls to the LEFT, so it's negative.
2) The 58N force pulling to the RIGHT (this is probably a rope or something.)

NET FORCE on small block is:
F2_net = 58N – friction

Now that we know the net FORCE on the two masses, we can write equations for their acceleration, using the famous "F=ma" (which we'll call "a=F/m", same thing):

Acceleration of slab:
a_slab = F1_net / m1
= friction / m1

Acceleration of block:
a_block = F2_net / m2
= (58N – friction) / m2

We're close, but we still don't know WHICH coefficient of friction to use, so we don't know how to calculate the friction. But here's a hint: We can look at the accelerations to guess whether the block & slab are sliding against each other or not. (Because, if the accelerations are DIFFERENT, they must be sliding. If they're the SAME, they must be moving together and therefore NOT sliding.)

So: Let's first pretend that a_block = a_slab (i.e., no sliding), and see if that leads to any contradictions:

If no sliding, then:
a_block = a_slab
(58N – friction) / m2 = friction / m1
58N/m2 – friction / m2 = friction / m1
58N/m2 = friction (1/m1 + 1/m2)
58N/9kg = friction (1/(21kg) – 1/(9kg))
6.444m/s² = friction (0.1587/kg)
friction = 6.444m/s² / (0.1587/kg) = 40.6N

Now, IF the block isn't sliding, the MAXIMUM possible force of friction is:

(normal force)(coeff. of static friction)
= m2(g)(0.30)
= (9kg)(9.8m/s²)(0.30)
= 26.46N

So, this is not enough! We saw above that we would need 40.6N of friction to make the two blocks "stick together". The static friction coefficient is not high enough to make that happen; and that means our initial assumption was wrong, and the blocks are sliding. That means we really must use the OTHER coefficient (0.20)

So the REAL force of friction is:
(normal force)(coeff. of KINETIC friction)
= m2(g)(0.20)
= (9kg)(9.8m/s²)(0.20)
= 17.64N

Finally, now that we've calculated the friction, plug that back into the equation for a_slab:

a_slab = friction / m1
= (17.64N) / (21kg)
= 0.84 m/s²

Answer 3

Hey
Ok, here goes:

The Force applied (Fapp) on the block is 58N, but in order for that force to move the block, it has to overcome the frictional force that exists between the block and the slab.
Therefore, we have to find the frictional force (Ff) first by using:
Ff = Fn(Ck), where Fn is the normal force of the block, and Ck is the coefficient of kinetic friction.
Note: we don't use the Coefficient of static friction because,
the coefficient of static friction anly exists when the two bodies are at rest, but since in our case, we intend to move them, we use Ck.
Fn ( normal force ) is the force that exist between the block and the gravitational field of the earth. In other words it is simply the weight of the block.
Therefore, Fn = Mb(Ge), where: Mb is the mass of the block and Ge is the gravitational force of the earth, which is approximately equal to 10m/s^2.
Therfore, by substituting the mass and the gravitatonal field we find the Fn.
Fn = 9kg(10m/s^2)
Fn = 90N.
We have found the Fn, now all we have to do now is to find the frictional force (Ff) that exists between the slab and the block.
We use :
Ff = Fn(Ck)
Therefore Ff = 90N(0.20)
Ff = 18N !
We have found the Ff and now all we need to do is apply this equation:
Fapp - Ff = Fused Where, Fused is the force needed to withstand the frictional force and move the body.
Therefore, Fused = 58N - 18N
Fused = 40N.
After the above, we only have to use a simple Equation to get at the answer.
I am sure you know this Equation:
F = ma, where m is mass and a is acceleration.
Since the question asks us for the acceleration, we use the above eq.
That is : a = F/m
Since we have already found the F to be 40N and the mass to be 9 kg:
a = 40N/9kg
a = 4.45m/s^2
Walla!!!!
The acceleration of the body is 4.45m/s^2.

P.S. I hope that's the right answer. If it is Holla at me and if it ain't well I gave it all I got.
Hope I helped
Alex

Answer 4

Sum of forces on slab:

T - mu (m1)g = m2 * a1

Sum of forces on block:
mu (m1)g = m1 * a2

Assume that they stick together (a1 = a2)

amax = mustatic * g

Tmax = mustatic g (m1+m2)

If T is smaller than that, then a = T / (m1+m2) and you're done.

If T is bigger than that, then you've got slippage:

T - mukinetic (m1)g = m2 * a1
mukinetic (m1)g = m1 * a2

Two equations and two unknowns--a1 and a2. Do the algebra to find a1.

Answer 5

Since there is no friction between the floor & slab, it's acceleration is infinite.