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Frictionless Ramp question, should be very simple

  • #1
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Homework Statement


A picture of the problem and the original question can be seen http://images3a.snapfish.com/232323232fp733;9>nu=52::>379>256>WSNRCG=335:73<377347nu0mrj".


Homework Equations





The Attempt at a Solution



Note that I made my coordinates so that Fn and T on m1 are the y and x axis respectively, also g will be rounded to 10 m/s^2 even though 9.8 is closer to the real value.

So first I know:

[tex] T=m_{2}g [/tex]

[tex] F_{n}=m_{1}gcos(30) [/tex]

So the only movement should be in the x direction so Fnet in y should be zero but here are my components:

[tex] F_{net_{x}}=-m_{1}gsin(30)+T=ma [/tex]

[tex]F_{net_{x}}=-m_{1}gsin(30)+m_{2}g=ma [/tex]

[tex] f_{net_{y}}=F_{n}-m_{1}gcos(30)=0 [/tex]

and i think I should solve Fnetx for a with m = 5, correct?

[tex] F_{net_{x}}=-(5)(10)(\frac{1}{2})+(7)(10)=(5)a [/tex]

[tex] F_{net_{x}}=-25+70 = 5a [/tex]

[tex] F_{net_{x}}= \frac {45}{5} = a [/tex]

[tex] a = 9 [/tex]

which is not the correct answer, where did I go wrong?
 
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Answers and Replies

  • #2
ehild
Homework Helper
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Try to use the more correct value g=9.8 m/s2.

ehild
 
  • #3
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Turns out I was suppose to use the combined mass of both objects for m, correct answer is 3.75 m/s^2.

I used 10 m/s^2 because we were told to, I would always use 9.8 if it were up to me :)
 
  • #4
ehild
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So first I know:

[tex] T=m_{2}g [/tex]


T would be m2 g only when m2 is in rest. But it accelerates downwards, with the unknown acceleration a.

ehild
 
  • #5
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how is the acceleration of m2 unknown? its a=g=-10 m/s^2, right? I don't get what your saying
 
  • #6
jhae2.718
Gold Member
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There are two forces on m2, tension T and weight m2g. Using Fnet=manet, the block has a net acceleration a that is dependent on gravity and tension. (Finding this net force is left as an exercise to the reader.) Tension is dependent on the other block since the rope is unstretchable.
 
  • #7
ehild
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how is the acceleration of m2 unknown? its a=g=-10 m/s^2, right? I don't get what your saying
NO, both masses move together as they are connected with a rope. The magnitude of the acceleration is a, for both of them.
If m2 accelerated with g downward, the tension would be zero.

ehild
 
  • #8
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Ok fair enough, but this had nothing to do with the problem either way though as the correct answer is 3.75 m/s^2 which you get by changing m to 12 instead of 5... This is truly introductory physics here.... like chapter 5 in the book. Nothing we have done in class or otherwise has talked about the hanging block having some acceleration that is not g.
 
  • #9
ehild
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Do you know what acceleration is?

ehild
 
  • #10
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I didn't see your post ehild before i replied. So this is a reply to you specifically,

if the block is just hanging there then it most certainly would have an acceleration -down- at 10 m/s^2 and the tension in the line for the hanging block would be 10 m/s^2 -up- so there would be no movement if the other block wasnt moving but none the less it would still have this force on it, no? and this tension force of 10 m/s^2 is translated into the x direction for m1 which is why this tension force is equal to m2*g which is why my equation worked...

I am not really trying to argue about how the laws of physics work here but my equation worked without taking any of this into account....(what your talking about)
 
  • #11
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acceleration is F/m if there is no force there is no acceleration, no acceleration no force.
 
  • #12
ehild
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What forces do act on m2?

ehild
 
  • #13
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The force of gravity (which is m2*g) tension (which is -m2*g) and the x component of gravity of the other block (which is m1*g*sin(30)), right? The last one is because they are connected so the x component of gravity of m1 should slow down the acceleration of m2.
 
  • #14
ehild
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The x component of gravity m1*g*sin(30) acts only on m1. It is the tension that acts on both boxes because they are connected. The tension will slow down the acceleration of m2.

ehild
 
  • #15
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so the x of gravity 'lowers' the tension force pulling m1 to the right which is the same as m2 down, put another way, the T force is m2*g-m1*g*sin(30)?
 
  • #16
ehild
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No. Look at your own very good free-body diagram.

What forces act on m2 according to it?

What forces act on m1?

ehild
 
  • #17
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m2: T, Fg

m1: T, Fg, Fn

but if the x component of gravity is not 'counter to' the tension pulling m1 to the right (so gravity to the left, vector wise) then I think i am more confused than when we started.....
 
  • #18
ehild
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The x component of gravity IS 'counter to' the tension but not equal to it.

The resultant forces are

on m2: (7 kg) m2 g acts downward, T upward. The resultant force is m2 g-T, is not it?
on m1 (5 kg) The normal component of gravity cancels out with the normal force. Gravity has a component of m1gsin30 downward along the slope and T upward, also parallel to the slope.
So the resultant force on m1 is T-m1gsin(30).

Both boxes move together with equal speeds, all the time. The acceleration is change of speed in unit time. The acceleration is also the same for both boxes.

Newton's second law states that ma=F(resultant).

Supposing that m1 moves upward on the slope and m2 moves vertically downward,

m1a = T-m1gsin(30),
m2a=m2g-T.

Adding up the equations, T cancels.

m1a +m2a=m2g-m1gsin(30).

Solve for a.

ehild
 
  • #19
you CANNOT equate forces on m2 as it is accelerating
 
  • #20
8
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use newtons 2nd law
both the masses are moving with same acceleration...by common sense...7 kg block is moving down.
lets consider forces acting on 5 kg block...
m(9.8)sin30
and
T...
as the 5 kg block is moving upward...t>m(9.8)sin30
therefore T-m(9.8)sin30=ma

do the same for block of mass 7kg



we get 2 equations-
T-5(9.8)sin 30 = 5a
7(9.8)-T=7a...solve these...
 

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