Newton's Second law problem

  • Thread starter Nemean
  • Start date
  • #1
Nemean
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 I am not sure what i am forgetting
 

Attachments

  • physics.png
    physics.png
    3.6 KB · Views: 405

Answers and Replies

  • #2
grzz
998
15
Are the masses at rest or moving?
 
  • #3
Nemean
12
0
i believe they are moving to the left
 
  • #4
grzz
998
15
Then there is motion and coefficient of KINETIC friction must be used.
 
  • #5
Nemean
12
0
If i use (.2)(200cos(30)) + (.2)(400sin(30)) only equals 74N and the answer is 63.8
 
  • #6
grzz
998
15
Correcting the coefficient of friction is the FIRST step.
Now you have to explain how you got the above expression.
 
  • #7
Nemean
12
0
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
grzz
998
15
Now i suggest you draw a diagram for M1 only with all the forces acting on M1.
 
  • #9
Nemean
12
0
O right, the applied force. Now would i add that to what i had gotten before? so 100cos(30) ?
 
  • #10
grzz
998
15
Why don't you use F(net) = Ma for M1 along the horizontal?
 
  • #11
Nemean
12
0
so wouldn't that be Fx = Facos(30) - T
 
  • #12
grzz
998
15
Fa = ?
 
  • #13
Nemean
12
0
force applied = 100N
 
  • #14
grzz
998
15
What about the friction on M1?
 
  • #15
Nemean
12
0
.2(200cos(30)?
 
  • #16
grzz
998
15
Not correct.
M1 is in vertical equilibrium. hence find the normal reaction on M1. THEN the friction force on M1.
 
  • #17
Nemean
12
0
so it would be .2(100cos(30))?
 
  • #18
grzz
998
15
use equilibrium in the vertical direction for M1.In the vertical dir there are THREE forces.
 
  • #19
Rayquesto
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
Nemean
12
0
Normal force = mg - Force applied y
(20*10) - 100sin(30) = 150N

Fg = mg
Fg = 200N
 
  • #21
grzz
998
15
Correct. Normal force for M1 is 150N.
NOW calculate the frictional force on M1.
 
  • #22
Nemean
12
0
it would be 15N
 
  • #23
grzz
998
15
System is moving . hence do not use static coeffienct.
 
  • #24
Nemean
12
0
then it would be 30
 
  • #25
grzz
998
15
correct. now you can use F = Ma for M1 downwards along the inclined plane.
 

Suggested for: Newton's Second law problem

  • Last Post
Replies
3
Views
334
Replies
9
Views
541
Replies
5
Views
522
  • Last Post
Replies
7
Views
245
  • Last Post
Replies
4
Views
420
Replies
14
Views
2K
Top