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Homework due online @ 9pm CST. 3 problems over Frictional forces.

  1. Oct 26, 2008 #1
    Homework due online @ 11pm CST. 3 problems over Frictional forces.

    ATTACHMENT TO HELP! it has diagrams and all. please this has taken much time to do, help :] I've used the format and rules in the sticky.

    Well guys, basically, I am an AP physics C student who needs some help with his homework. I am not alone, seeing as I am with my friends doing this homework together. The homework is due online where you put the answers into the computer, yet out of 16 problems, we were unable to do 3(5 parts total) of them. We've made many attempts, so i'm just hoping you guys can help! Its due tonight :/ I have attached a PDF of the homework so you may look @ the diagrams of the problems. The numbers of the problems are #2 ; 11&12 ; 13&14

    Problem 2 attachment!
    1. Three blocks in contact with each other are pushed across a rough horizontal surface by a 86 N Force as shown.(Look at PDF please) If the coefficient of kinetic friction between the blocks and the surface is .15, find the magnitude of the force exerted on the 7.8 kg block by the 11 kg block. Answers in units of N




    2. Relevant equations

    Frictional force = Normal force x coefficient of friction
    w = mg
    f = ma (yes basic i know!)

    3. The attempt at a solution

    Well, we tried combining the first two masses to get 9.8 kg, and we found the netforce on it by subtracting the total frictional force from 86 N, which gives us 71.594 N. This gave us the force pushing forward onto the 11 kg block, but we need to find the force that the 11 kg block exerts upon the 7 kg box. We then found the frictional force upon the 11 kg box, which is causing the box to push back. we added the two forces which gave us 87.764, which was wrong in the computer. HELP PLEASE!

    Problem 11&12 please look at attachment!
    1. A block of mass .9 kg rests on the inclined surface of a wedge of mass 1.9 kg. The wedge is acted upon by a horizontal force F and slides on a frictionless surface. If the coefficient of static friction between the wedge and the block is .8 and the angle of the incline if 33degrees, find the minimum value of F for which the block does not slip. Also, find the maximum value of F for which the block does not slip. Answer in units of N

    2. Relevant equations

    f= ma
    Ff = u Fn
    w = mg

    3. The attempt at a solution

    We found out the frictional force needed to keep the block still, but we cannot figure out how to incorporate force F into anything in this problem. It is extremely confusing and is hard to solve. STUMPED! all 3 of us. help :/.

    mg sinX = the force the block will slide down with and = Ff needed to keep the block still.

    Problem 12 & 13 attachment please!!!

    1. An 11 kg block rests on a 7 kg bracket shown in the figure. The 7kg bracket sits on a frictionless surface. The coeff. of friction b/w the bracket and the 11 kg block are .4 static and .2 kinetic. what is the maximum acceleration(m/s^2) of the bracket if the 11 kg block is not to slide on the bracket? Also, what is the corresponding force applied? Answer in units of N

    2. Equations associated

    f=ma
    Ff = uFn
    w = mg
    extrapolation!

    3. Attempts!

    somehow, we got this wrong. We decided that for the 11 kg block to not move, the force pulling the block must not overcome the static friction.

    w = mg which is also Fn yields Ff = (Fn)(.4) which gave us an answer of 43.12 N of static frictional force. We assumed that this means the force may not overcome 43.12 N, so the force of the pulling rope was 43.12 N, and the total mass of the system was 18 kg, on a frictionless surface.
    We did F=ma on the system, leading it to be 43.12 = (18)(A); we got acceleration to be 2.39555, but it was wrong when placed into the computer, help please!!!!!!!!
     

    Attached Files:

    Last edited: Oct 26, 2008
  2. jcsd
  3. Oct 26, 2008 #2

    Hootenanny

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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to Physics Forums.

    Let's take these problems one at a time.
    You should have realised that this was wrong before you entered it into the computer, how can the force exerted between the blocks possibly be greater than the force that is applied?

    Anyway the key the point to realise here is that for the blocks to remain in contact, they each must accelerate at the same rate.
     
  4. Oct 26, 2008 #3
    thank you for the welcome :]

    i realize my friends and I have made stupid mistakes, but a good thing is that the homework isn't due til 11 pm CST lol.

    I don't know where to go from anything on this problem. As easy as the problem may seem compared to the other problems on the assignment, this stumps me. If the acceleration of all 3 blocks is constant, where do i go from there? I don't understand how to find the force of the 11 kg block pushing back onto the 7kg block. Ff? w? blah :/
     
  5. Oct 26, 2008 #4
    we figured them out thanks to other friends.

    thanks anyways! solved!
     
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