# Homework Help: Newton's Second law problem

1. Oct 2, 2011

### Nemean

1. The problem statement, all variables and given/known data
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

2. Relevant equations

F = ma
FR = μmg

3. The attempt at a solution

$\sum$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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2. Oct 2, 2011

### grzz

Are the masses at rest or moving?

3. Oct 2, 2011

### Nemean

i believe they are moving to the left

4. Oct 2, 2011

### grzz

Then there is motion and coefficient of KINETIC friction must be used.

5. Oct 2, 2011

### Nemean

If i use (.2)(200cos(30)) + (.2)(400sin(30)) only equals 74N and the answer is 63.8

6. Oct 2, 2011

### grzz

Correcting the coefficient of friction is the FIRST step.
Now you have to explain how you got the above expression.

7. Oct 2, 2011

### Nemean

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. Oct 2, 2011

### grzz

Now i suggest you draw a diagram for M1 only with all the forces acting on M1.

9. Oct 2, 2011

### Nemean

O right, the applied force. Now would i add that to what i had gotten before? so 100cos(30) ?

10. Oct 2, 2011

### grzz

Why don't you use F(net) = Ma for M1 along the horizontal?

11. Oct 2, 2011

### Nemean

so wouldnt that be Fx = Facos(30) - T

12. Oct 2, 2011

### grzz

Fa = ?

13. Oct 2, 2011

### Nemean

force applied = 100N

14. Oct 2, 2011

### grzz

What about the friction on M1?

15. Oct 2, 2011

### Nemean

.2(200cos(30)?

16. Oct 2, 2011

### grzz

Not correct.
M1 is in vertical equilibrium. hence find the normal reaction on M1. THEN the friction force on M1.

17. Oct 2, 2011

### Nemean

so it would be .2(100cos(30))?

18. Oct 2, 2011

### grzz

use equilibrium in the vertical direction for M1.In the vertical dir there are THREE forces.

19. Oct 2, 2011

### Rayquesto

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. Oct 2, 2011

### Nemean

Normal force = mg - Force applied y
(20*10) - 100sin(30) = 150N

Fg = mg
Fg = 200N

21. Oct 3, 2011

### grzz

Correct. Normal force for M1 is 150N.
NOW calculate the frictional force on M1.

22. Oct 3, 2011

### Nemean

it would be 15N

23. Oct 3, 2011

### grzz

System is moving . hence do not use static coeffienct.

24. Oct 3, 2011

### Nemean

then it would be 30

25. Oct 3, 2011

### grzz

correct. now you can use F = Ma for M1 downwards along the inclined plane.