# Newtons 3rd Law questions.

1. Oct 16, 2006

I am needing help on a few problems that I have to do for class. If anyone could help on any of them that would be great.

1)Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s. The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 m.
a. What constant force must Bob exert on the rock to throw it with this speed?
b. If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

2) a block of mass rests on a 20 deg slope. The block has coefficients of friction 0.8 (static) and 0.5 (kinetic) with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.
a. What is the minimum mass that will stick and not slip?
b. If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

3) The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force causes both blocks to cross a distance of 5.0 m, starting from rest.
a. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

I have already tried doing these problems many times each and am running out of chences to be wrong so I am just hoping someone out there who is better in physics can come up with the right way to do it! Thanks!

Last edited: Oct 16, 2006
2. Oct 16, 2006

### kcirick

Can you post what you've done so far? For all of these questions, drawing diagrams is a must, along with free-body diagram.

3. Oct 16, 2006

Okay...here is what I have tried so far:

1)I tried getting the a by using v^2=v_0^2+2ax
this gave me a=450m/s^2 which I assumed was wrong
My thinking was that if I found a I could just plug it into F=m*a using the combined masses. I knew that was wrong and I pretty much hit a road block with this one.

2) For the 2kg mass: T_2-m_2*g=0
For unknown mass: y: N-mg+T_1sin(20)=0
x: T_1cos20=f_s=N*mu_s=m_1gcos20*mu_s
then I combined the equations and got
m_1*g-mu_s*m_2*g*cos20-m_1*g*sin20=0
I solved for m_1 and got 2.19...which is wrong.

Since I never got a I didnt try b yet.

3)a_top=a_bot=a
For the top block: sumF_x=F=m_t*a+f_s
F=m_t*a+mu_s*m_t*g
For the bottom block: sumF_x=-f_s-f_k=-mu_s*m_t*g-mu_k*(m_t+m_b)*g
I added them and set them equal to zero and got
m_t*a-mu_k*(m_t+m_b)*g=0
a=8.575
then plugged that into: 5=(1/2)at^2 and got t=1.37 sec

4. Oct 16, 2006

### gunblaze

Yup. Thats the value you should be getting for ur acceleration. However, you do not multiply by the combined masses. You should just multiply it with the mass of the stone which is .5kg.

For b, what you need to use is to plug the values you got into the formula of the conservation of momentum.

For Qn 2, You have forgotten about ur static friction. It allows you to further decrease the mass of m_1 without it slipping.

Last edited: Oct 16, 2006
5. Oct 16, 2006

### gunblaze

oops. I'm sorry. Eh, for b, what u could do is to just take the reaction force by the rock back on Bob divided by his weight. Basically, if the surface is frictionless, and provided there is no air resistance, there will not be a constant recoil velocity, because Bob will just keep accelerating in the opposite direction untill there is some resistive force which can cancel his reaction force to allow him to travel at constant speed.

6. Oct 16, 2006

Thanks for the help with 1 gunblaze!

7. Oct 16, 2006

### gunblaze

8. Oct 16, 2006

I attached the file for question 3.

#### Attached Files:

• ###### 3image.doc
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9. Oct 16, 2006

Here is also the image for 2.

#### Attached Files:

• ###### 2image.doc
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53.5 KB
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10. Oct 17, 2006

let me know if there is a problem seeing any of the images.

11. Oct 17, 2006

Any suggestions? Could certainly use some help from those out there much smarter in this subject than I

12. Oct 17, 2006

Regarding 2. As you did, use the equation of equilibrium for the 2 kg mass to find the tension in the rope. Now draw a free body diagram for the unknown mass m, which must be at rest. The forces acting at it are the static friction force, the tension (in the same direction), the weight, and the normal force (reaction from the ground on the object). Now, set a local coordinate system up so that the x axis is in the direction of the incline. Write down the equations od equilibrium for both x and y directions. You should be able to obtain m easily.

13. Oct 17, 2006

Thanks for the help with number 2 radou. Anyone have any advice for number 3?

14. Oct 17, 2006

### gunblaze

Alright, for question 3, You need to know what really is the thing that they want you to find. What they really want to know is the max force allowable to allow ur 2 blocks to move 5m without them seperating when the force is only applied to the top block. Instead of making life easier, they asked for time which can be found easily if the max force is found.

15. Oct 17, 2006

Well I keep trying to go about it by solving for acceleration but that isnt getting me anywhere. So what I have is:

Top: F_x=m_top*a+mu_s*m_top*g
Bottom: F_x=-mu_s*(m_top+m_bot)*g

then do I set them equal and solve for a?

Then I have been trying to plug a into the eqn
x=x_0+v_0*+1/2at^2

How do I use any of this to get the max force?

16. Oct 17, 2006

### gunblaze

Alright, let's do this step by step. Firstly, how can we get the max force? Whats the max force that can be applied to the top box before the top box actually slides off? Consider ur friction. The max force allowable simply cannot exceed the force of friction between the 2 blocks right?

17. Oct 17, 2006

So the max force would be f_s=mu_s*N. So do you use the N of the whol system so N=(m_top+m_bot)*g or just the top block N=m_top*g?

So when we get the F we just set it equal to F=ma and solve for a? Do I use the combined masses for m?

Was I right originally in saying that I plug the a into the kinematic eqn to get 5=1/2at^2 and solve for t that way?

18. Oct 17, 2006

### gunblaze

ah. Finally, we back on the right track.. Yes, the normal force that will be causing the friction between the 2 blocks will just be the weight of the top block.

Note, when you are finding the friction between the ground and the top block, you also dun take into consideration the weight of the floor right?

When you find the max force allowed, do not forget, you do have to consider also the force of friction between the whole system and the ground too.

Yes, the last statement is right, you do need to put in the a value into that kinematic equation to find time taken.

19. Oct 17, 2006

So for the top F=mu_s*m_top*g
and for the bottom F=(-mu_s*(m_top+m_bot)*g)-(mu_k*(m_top+m_bot)*g)?

I am still pretty confused about the equations I am supposed to use. Physics is pretty much my worst subject so I am pretty clueless most of the time.

20. Oct 17, 2006

### gunblaze

Yup. That's about it. What you are actually trying to find for the top eqn is the frictional force between the blocks while for the bottom eqn, you are actually finding the frictional force between the 2 blocks and the ground. See the difference?

The reason for finding the frictional force between the 2 blocks is to find the max force allowed to prevent sliding.. While for the frictional force between the ground and the 2 blocks, what you are actually trying to find is the resultant force on the 2 blocks that make them move forward together.