anonymous
  • anonymous
Please help http://oi67.tinypic.com/2mgozms.jpg
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@matt101
anonymous
  • anonymous
@ganeshie8
wmj259
  • wmj259
maximum magnitude a static friction force between the box and truck bed can have is f(max)=mui*normal normal is basically mg But you also have to use newtons laws on this problem. F=ma ^ in this case you set the force equal to the frictional maximum force. So thats then mui(mg)=ma If you solve for acceleration, the masses cancel out. So you have left: a=mui*g

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wmj259
  • wmj259
in this case your mui due to static friction is 0.51
wmj259
  • wmj259
I hope this helped.
anonymous
  • anonymous
Okay so (.51N)(9.81*351)
wmj259
  • wmj259
No the mass gets canceled out because of the newtons formula.
anonymous
  • anonymous
Okay so just (.51N * 9.81)?
wmj259
  • wmj259
thats not a N, thats mui.
wmj259
  • wmj259
http://www.mrwaynesclass.com/ap/Ch4_freebody/CH04W59.pdf. Look at how this page derives the equation.
anonymous
  • anonymous
404 error
anonymous
  • anonymous
So you wouldn't find acceleration by multiplying the gravity by mui?
anonymous
  • anonymous
@AlexandervonHumboldt2
wmj259
  • wmj259
Wheres the error404 coming from :)
anonymous
  • anonymous
It did that when I clicked on it.
anonymous
  • anonymous
@aaronq
wmj259
  • wmj259
http://www.mrwaynesclass.com/ap/Ch4_freebody/CH04W59.pdf I think the extra spaced messed up the link.
wmj259
  • wmj259
just copy and paste the link withouth the extra spaces.
anonymous
  • anonymous
@Compassionate
anonymous
  • anonymous
For this would I just take (.51)(9.81)=5.0031
anonymous
  • anonymous
Always start with: |dw:1447296703167:dw| From there, you know that Force of gravity is equal to the normal force, since the object will only move horizontally. So the y-components of force will be equal to zero: Fnormal - Fgravity = 0 N -->eqn 1 To solve for the Friction Force which is required to solve for the maximum acceleration, you need the formula for the normal force. From the eqn 1, rearrange it in terms of Fnormal: Fnormal = Fgravity Fnormal = mass x acceleration due to gravity (SO CALCULATE THIS FIRST- just leave the mass as variable "m" for now) The formula for the Friction force is: |dw:1447297058064:dw| where Fn is the normal force, µ is the coefficient of friction Subsitute the values and solve for Ff. NEXT SOLVE FOR ACCELERATION: According to Newton's second law, the object must move on the direction of the friction force, since it is the only force remaining that will cause the object to move: Ff = mass x acceleration rearrange the formula and you'll get : acceleration = force / mass variable 'm' should cancel out.
anonymous
  • anonymous
that is for letter A.
anonymous
  • anonymous
For letter B, you have to refer to your solution on letter A. Since you cancel the 'm' or the mass on your calculation, do you think having different mass matters to have a safe acceleration?
anonymous
  • anonymous
So from letter A, what normal force did you get?
anonymous
  • anonymous
Fgrav= (9.81)(351) = 3443.31 =Fnorm
anonymous
  • anonymous
right, now solve for the friction force using that normal force and the coefficient of friction, what will you get?
anonymous
  • anonymous
So would the m be .51?
anonymous
  • anonymous
it is µ= 0.51, not the m. just ignore the 'm' for now--we'll use it on letter B.
anonymous
  • anonymous
So Ff = mui 3443.31
anonymous
  • anonymous
yeah but mui is 0.51 , so you can solve for Ff: 0.51 x 3443.31
anonymous
  • anonymous
Alright I was thinking that. I got Ff=1756.0881
anonymous
  • anonymous
okay good :D now solve for acceleration
anonymous
  • anonymous
I divide the 1756 by the mass right
anonymous
  • anonymous
yep
anonymous
  • anonymous
..and you'll get?
anonymous
  • anonymous
5.0031
anonymous
  • anonymous
yes :) that will be your acceleration, don't forget your unit :)
anonymous
  • anonymous
Right. 5.00 m/s^2
anonymous
  • anonymous
yes, perfect. For B, instead of substituting the values, use variables instead to solve for 'a' ... and go back the question I asked about letter B.
anonymous
  • anonymous
I don't really see why the mass of the truck would matter.
anonymous
  • anonymous
yes, your answer is kinda right, but you have to know the reason why the mass of the loads would not matter :)
anonymous
  • anonymous
|dw:1447299955473:dw|
anonymous
  • anonymous
So it because the mass would just cancel out
anonymous
  • anonymous
yeah , that's how I will reason it :P
anonymous
  • anonymous
Makes sense.
anonymous
  • anonymous
Thanks for the help
anonymous
  • anonymous
np :) goodluck with the rest of your physics
anonymous
  • anonymous
Thanks

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