Friction Force Homework Problem

[Tensaiga]

Hello, i have been asking a lot of question lately, i had 60 questions to do for hw, and I'm kinda confused on some of them...

A 0.5kg block is placed on top of a 1.0kg block, the coefficient of static friction between the two blocks is 0.35. The cofficient of kinetic friction between the lower block and the lever table is 0.20. What is the maxinimum horizontal force that can be applied to the lower block without the upper block slipping?

Some Thoughts:
First i found out the kinetic frictional force,between the lower block and the table. (i don't know if you should add the two masses together, since block 2 is on top of block 1, so does it change block 1's normal force? anyways i didn't add 0.5kg.)

Block 1: mg(us) = (0.5kg)(9.8N/kg)(0.35) = 1.715N
Block 2: mg(uk) = (1.0kg)(9.8N/kg)(0.20) = 1.96N

when Force applied = Frictional, no accleration.
So, F applied must first overcome block 2's frictional force, then you have a net force, but that force must not be greater than 1.715N

So i added the number forces: 1.96N + 1.715N = 3.675N. The force applied must be less than this... but i was wrong, our book says the correct answer is 8.1N... i wonder how they got that.

Thanks.

 

[Andrew Mason]

Hello, i have been asking a lot of question lately, i had 60 questions to do for hw, and I'm kinda confused on some of them...

A 0.5kg block is placed on top of a 1.0kg block, the coefficient of static friction between the two blocks is 0.35. The cofficient of kinetic friction between the lower block and the lever table is 0.20. What is the maxinimum horizontal force that can be applied to the lower block without the upper block slipping?

Some Thoughts:
First i found out the kinetic frictional force,between the lower block and the table. (i don't know if you should add the two masses together, since block 2 is on top of block 1, so does it change block 1's normal force? anyways i didn't add 0.5kg.)

Block 1: mg(us) = (0.5kg)(9.8N/kg)(0.35) = 1.715N
Block 2: mg(uk) = (1.0kg)(9.8N/kg)(0.20) = 1.96N

when Force applied = Frictional, no accleration.
So, F applied must first overcome block 2's frictional force, then you have a net force, but that force must not be greater than 1.715N

So i added the number forces: 1.96N + 1.715N = 3.675N. The force applied must be less than this... but i was wrong, our book says the correct answer is 8.1N... i wonder how they got that.

If you apply enough force to get the blocks moving at some small speed (no acceleration), what is the horizontal force that the lower block exerts on the upper block? What is it if you apply a greater force?

Draw a free body diagram showing the horizontal forces acting on the upper and lower blocks. Set the force on the lower block to be just a bit greater than the force of static friction between the upper and lower block. What happens?

Also, you have to add the weight of the two blocks for the normal force on the table surface (to determine the force of kinetic friction between the table and the lower block).

AM

 

[Tensaiga]

Sorry i forget to sent the diagarm along with it.

what is the horizontal force that the lower block exerts on the upper block?

What do you mean by that? they are perpendicular, shouldn't it have no effect on each other?

 

[Andrew Mason]

Sorry i forget to sent the diagarm along with it.

what is the horizontal force that the lower block exerts on the upper block?

What do you mean by that? they are perpendicular, shouldn't it have no effect on each other?

I meant: when you apply a horizontal force to the lower block, what is the horizontal force that the lower block exerts on the upper block?

AM

 

[Tensaiga]

umm, the same? as the lower block?

 

[Doc Al]

I will ask the question differently. First question: What is the means by which the bottom block is able to exert a horizontal force on the top block? (What kind of force is it?)

 

[Andrew Mason]

umm, the same? as the lower block?

No. Think of it this way: Assume that the force applied to the lower block is greater than the force of kinetic friction. What happens to the lower block? What is the horizontal force on the upper block from the lower block if they both move together? (hint: Newton's second law of motion applied to the upper block).

AM

 

[Tensaiga]

oh wait, i get it now, well i got the right answer but i don't know why i did that, could someone explain to me why i did that??

OK: First the Horizontal Force required to start the movenment, that's
(1.0kg+0.5kg)*(9.8)*(0.20) = 2.94N

Then in order for the block to fall off, it must over come the static friction between the two blocks, which is (1.0 + 0.5)*(9.8)*(0.35) = 5.145N

Than you add the two together that's should be the maxinimum force allowed without the blocks slipping...BUT what i don't understand is that why do i have to add the mass of the lower block AND the upper block together to get the force of static friction? isn't it only suppose to be (0.5)*(9.8)*(0.35), without the lower block? I'm confused on this part.

PS: yea i understand the horizontal part now, but not static...
THANKS

 

[Beam me down]

BUT what i don't understand is that why do i have to add the mass of the lower block AND the upper block together to get the force of static friction? isn't it only suppose to be (0.5)*(9.8)*(0.35), without the lower block? I'm confused on this part.

PS: yea i understand the horizontal part now, but not static...
THANKS

You want to start the system of the two blocks moving. The normal force for the two blocks is proportional to the sum of their masses. The acceleration of the two blocks also needs to consider both their masses. When you consider the maximum force before the top block moves relative to the bottom block, then you consider the masses seperately.

 

[Andrew Mason]

oh wait, i get it now, well i got the right answer but i don't know why i did that, could someone explain to me why i did that??

OK: First the Horizontal Force required to start the movenment, that's
(1.0kg+0.5kg)*(9.8)*(0.20) = 2.94N

Then in order for the block to fall off, it must over come the static friction between the two blocks, which is (1.0 + 0.5)*(9.8)*(0.35) = 5.145N

Than you add the two together that's should be the maxinimum force allowed without the blocks slipping...BUT what i don't understand is that why do i have to add the mass of the lower block AND the upper block together to get the force of static friction? isn't it only suppose to be (0.5)*(9.8)*(0.35), without the lower block? I'm confused on this part.

PS: yea i understand the horizontal part now, but not static...
THANKS

The static friction force is the MAXIMUM force that the lower block can exert on the upper block (horizontally). So let the force on the lower block be such that the two blocks stay together:

$$F = (m+M)a + \mu_k(m+M)g$$

(1)$$a = \frac{F}{m+M} - \mu_kg$$

Since the static friction force has to supply the accelerating force on the upper mass: [itex]ma < \mu_s mg[/tex]

So putting that condition into (1):

$$\mu_smg > ma = \frac{Fm}{m+M} - \mu_kmg$$

So:

$$F < (M+m)(\mu_sg + \mu_kg)$$

$$F < 1.5*(.35 + .2)*9.8 = 8.1$$N.

AM

 

[Tensaiga]

i see i see, thank you, the way you did that seems simple. Yet complicated, with all the variables, but i did understood that,thanks.