# Lack of a teacher

1. Sep 17, 2006

### rwofford

My teacher gave me this problem but we haven't even gone over two body probles with tension...especially those where one object is going another direction...
1) The drawing shows a large cube (mass = 21 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 2.1 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

I dont know where to begin...are there any sites that would be helpful to someone who has not been taught how to do these types of problems...?

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2. Sep 17, 2006

### Staff: Mentor

What is the magnitude of the force downward on the small front cube? What is the magnitude of the retarding frictional force upward due to the acceleration P?

3. Sep 17, 2006

### rwofford

isnt the acceleration of the large and small cube the same..? but they dont tell me the acceleration...

4. Sep 17, 2006

### Staff: Mentor

You're supposed to calculate the acceleration P where the front cube (yes, accelerating at P also) has its gravitational force down and its frictional force up balancing. Write the two equations for those two forces, and set them equal and see if that solves P for you. Let's see the work.

5. Sep 17, 2006

### Staff: Mentor

Gotta go. Good luck!

6. Sep 17, 2006

### Staff: Mentor

One has to calculate the horizontal force P which produces an acceleration.

P acts on the large and small cubes. The objective is determine the acceleration for which friction between small and large cube will prevent the small cube from falling.

So gravity pulls down the small cube, and the friction force operates vertically (in opposition to gravity). The friction force is simply the static coefficient * normal force which is simply the mass * acceleration.

The force P accelerates the large block against the small block, which also must be accelerating at the same rate for both to move together. But remember, mass resists acceleration or change in momentum.

Last edited: Sep 18, 2006
7. Sep 17, 2006

### rwofford

Fg=mg
=21(9.8)
=205.8 N

=(2.1)(9.8)
=20.58

so i know: Ns * Fn = ma
0.71*Fn=205.8N(a)

8. Sep 17, 2006

### rwofford

i might as well give up...i am horrible at physics

9. Sep 17, 2006

### Staff: Mentor

Hint 1: What is the only force supporting the weight of the small cube? (Astronuc gave you this one.)

Hint 2: Use that fact to calculate the required normal force between the two cubes.

Hint 3: Use that fact to determine the resulting acceleration of the small cube.

That's enough hints for now! Solve it algebraically (using symbols) as much as possible--don't be in a rush to plug in numbers.

10. Sep 17, 2006

### rwofford

the normal force=mg
so : 2.1(9.8)=20.58 N
so : 0.71(20.58)=ma
so : 14.6118=ma

11. Sep 17, 2006

### Staff: Mentor

No, the friction force equals the weight.

12. Sep 17, 2006

### rwofford

do i plug in 2.1 or 21 for m?

13. Sep 17, 2006

### Gokul43201

Staff Emeritus
Any effort is wasted unless you begin by drawing the free body diagrams for all the objects in the problem.

14. Sep 17, 2006

### Staff: Mentor

We are talking about the smaller cube, so use its mass.

Gokul speaks truth--do as he advises.

15. Sep 17, 2006

### rwofford

so : 14.6118 = 2.1a
so : 6.958 = a

16. Sep 17, 2006

### rwofford

i think ill give up now...i am dumb.....................................................................................................................................................................

17. Sep 17, 2006

### rwofford

is this right/???

18. Sep 17, 2006

### Gokul43201

Staff Emeritus
No, it is not.

You need to start at the beginning. Follow these steps EXACTLY:

1. Draw and label the forces acting on the little block.

2. Write down the net force (sum of all forces) acting in the vertical direction, and the net force acting in the horizontal direction. Use symbols to represent various quantities. Save the actual numbers for the very end.

3. Write down Newton's Second Law for each of these two directions, and plug in the expressions for the net force and acceleration in each direction.

Do these steps here, and we'll see how it goes....

19. Sep 18, 2006

### Staff: Mentor

How does one use the 'coefficient of friction', $\mu$? What is the relationship between friction and normal force?

This part is correct - the weight (or gravitational force) of the small block.
=(2.1)(9.8)
=20.58 N (One should write units)

Now one must determine the vertical friction force, which is proportional to the normal (horizontal) force, which is proportional to the acceleration of the small block.

That acceleration comes from the force P acting on both blocks.

I recommend doing as Gokul suggested with the force diagrams, and please refer to the steps mentioned by Doc Al.

Last edited: Sep 18, 2006