What is the Net Force on a System with Multiple Objects and Inclined Planes?

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Homework Help Overview

The discussion revolves around calculating the net force on a system involving multiple objects on inclined planes, with specific forces acting on each object, including normal and frictional forces.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the conditions of motion for the objects, questioning whether they are at rest or moving, and discuss the appropriate coefficients of friction to use. There are attempts to derive expressions for net force based on the components of forces acting on the objects.

Discussion Status

The conversation is active, with participants providing guidance on how to approach the problem, including suggestions to draw diagrams and clarify the forces acting on the objects. There is a focus on ensuring the correct application of kinetic friction due to the motion of the system.

Contextual Notes

Participants note the importance of distinguishing between static and kinetic friction, as well as the need to consider vertical equilibrium when calculating normal forces. There are references to specific values for forces and coefficients, which may be subject to verification.

Nemean
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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 I am not sure what i am forgetting
 

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Are the masses at rest or moving?
 
i believe they are moving to the left
 
Then there is motion and coefficient of KINETIC friction must be used.
 
If i use (.2)(200cos(30)) + (.2)(400sin(30)) only equals 74N and the answer is 63.8
 
Correcting the coefficient of friction is the FIRST step.
Now you have to explain how you got the above expression.
 
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
 
Now i suggest you draw a diagram for M1 only with all the forces acting on M1.
 
O right, the applied force. Now would i add that to what i had gotten before? so 100cos(30) ?
 
  • #10
Why don't you use F(net) = Ma for M1 along the horizontal?
 
  • #11
so wouldn't that be Fx = Facos(30) - T
 
  • #12
Fa = ?
 
  • #13
force applied = 100N
 
  • #14
What about the friction on M1?
 
  • #15
.2(200cos(30)?
 
  • #16
Not correct.
M1 is in vertical equilibrium. hence find the normal reaction on M1. THEN the friction force on M1.
 
  • #17
so it would be .2(100cos(30))?
 
  • #18
use equilibrium in the vertical direction for M1.In the vertical dir there are THREE forces.
 
  • #19
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
Normal force = mg - Force applied y
(20*10) - 100sin(30) = 150N

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

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