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Newton's Second law problem

  • Thread starter Nemean
  • Start date
  • #1
12
0

Homework Statement


Find the Net force on the system
see thumbnail for given.

Normal force action on object 1 - 150N
Normal force action on object 2 - 346N
Force of friction action on object #1 - 15N
Force of friction action on object #2 - 34.6N
M1 moves to the left

Homework Equations



F = ma
FR = μmg

The Attempt at a Solution



[itex]\sum[/itex]Fx = (.1)(20)(10)cos(30) + (.1)(40)(10)sin(30)
= 37.32N

The answer should be 63.8N im not sure what i am forgetting
 

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Answers and Replies

  • #2
993
13
Are the masses at rest or moving?
 
  • #3
12
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i believe they are moving to the left
 
  • #4
993
13
Then there is motion and coefficient of KINETIC friction must be used.
 
  • #5
12
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If i use (.2)(200cos(30)) + (.2)(400sin(30)) only equals 74N and the answer is 63.8
 
  • #6
993
13
Correcting the coefficient of friction is the FIRST step.
Now you have to explain how you got the above expression.
 
  • #7
12
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well that's were i get a bit lost. I need to find the net force, and i figured that if i used the x component of box 1 and box 2 so i got 200cos30 for box one, and 400sin30 for box two
 
  • #8
993
13
Now i suggest you draw a diagram for M1 only with all the forces acting on M1.
 
  • #9
12
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O right, the applied force. Now would i add that to what i had gotten before? so 100cos(30) ?
 
  • #10
993
13
Why don't you use F(net) = Ma for M1 along the horizontal?
 
  • #11
12
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so wouldnt that be Fx = Facos(30) - T
 
  • #12
993
13
Fa = ?
 
  • #13
12
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force applied = 100N
 
  • #14
993
13
What about the friction on M1?
 
  • #15
12
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.2(200cos(30)?
 
  • #16
993
13
Not correct.
M1 is in vertical equilibrium. hence find the normal reaction on M1. THEN the friction force on M1.
 
  • #17
12
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so it would be .2(100cos(30))?
 
  • #18
993
13
use equilibrium in the vertical direction for M1.In the vertical dir there are THREE forces.
 
  • #19
318
0
The way i would approach this problem is to first figure out how to find the force pulling at that certain angle using the coefficient of static friction since static means the forces are all in equilibrium and you can make the sum equal to zero. SO, you know that just before a netforce higher than zero is provided, the coefficient of static friction times the normal force is the friction.
 
  • #20
12
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Normal force = mg - Force applied y
(20*10) - 100sin(30) = 150N

Fg = mg
Fg = 200N
 
  • #21
993
13
Correct. Normal force for M1 is 150N.
NOW calculate the frictional force on M1.
 
  • #22
12
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it would be 15N
 
  • #23
993
13
System is moving . hence do not use static coeffienct.
 
  • #24
12
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then it would be 30
 
  • #25
993
13
correct. now you can use F = Ma for M1 downwards along the inclined plane.
 

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