# Homework Help: A box of mass 8.0kg is being dragged along the ground by a force of 30N.

1. May 14, 2008

### looi76

[SOLVED] A box of mass 8.0kg is being dragged along the ground by a force of 30N.

1. The problem statement, all variables and given/known data
A box of mass $$8.0kg$$ is being dragged along the ground by a force of $$30N$$.
(a) If the friction force is $$26N$$ what is the resulting acceleration?
(b) If the frictional force is one-quarter of the normal reaction force, what is the acceleration?

2. Relevant equations
F = m.a

3. The attempt at a solution

(a) $$m = 8.0kg \ , \ F = 30N$$

$$F = 30 - 26 = 4N$$

$$F = m.a$$

$$a = \frac{F}{m} = \frac{4}{8} = 0.5ms^{-2}$$

(b) $$m = 8.0kg , \ , F = 30 \times \frac{3}{4} = 22.5N$$

$$a = \frac{F}{m} = \frac{22.5}{8.0} = 2.8ms^{-2}$$

2. May 14, 2008

### Hootenanny

Staff Emeritus
Correct
Careful here, notice what the question says:

3. May 14, 2008

### looi76

Thnx Hootenanny,

Normal Reaction Force $$= 4N$$

$$m = 8.0kg \ , \ F = 4 \times \frac{3}{4} = 3N$$

$$F = m.a$$

$$a = \frac{F}{m} = \frac{3}{8} = 0.4 ms^{-2}$$

I don't thinks its correct, can you please explain?

By Normal reaction force, it means the reaction without the friction?

4. May 14, 2008

### Hootenanny

Staff Emeritus
No, the http://en.wikipedia.org/wiki/Normal_force" [Broken] is the force which acts on the box and is normal to the surface. It is the force which prevents the box from entering the surface.

Last edited by a moderator: May 3, 2017
5. May 14, 2008

### looi76

I understood what normal reaction force means, but tell now I don't know how to calculate it. Can you please show me an example or explain?

Thanks...

6. May 14, 2008

### Hootenanny

Staff Emeritus
Consider your example, what are all the forcing acting in the vertical plane?

7. May 14, 2008

### looi76

Forces acting on the vertical plane are 30N and 26N

8. May 14, 2008

### Hootenanny

Staff Emeritus
No, they are the forces acting on the horizontal plane. What are the forces acting in the vertical plane, i.e. in the 'up and down' direction?

9. May 14, 2008

### looi76

$$F = m.a$$

$$m = 8.0kg \ , \ a = g(gravity) = 9.81ms^{-2}$$

$$F = 8.0 \times 9.81$$

$$F = 78.5N$$

So, the normal reaction force is $$78.5N$$ ?

10. May 14, 2008

### Hootenanny

Staff Emeritus
Correct

Personally I would leave it as 78.48 N to prevent rounding errors later on.

11. May 14, 2008

### looi76

Thnx Hootenanny, So answer (b) would be:

$$F = m.a$$

$$F = 30 - \left(\frac{1}{4} \times 78.48\right) = 10.38N \ , \ m = 8.0kg$$

$$a = \frac{F}{m} = \frac{10.38}{8.0} = 1.3ms^{-2}$$

correct?

12. May 14, 2008

### Hootenanny

Staff Emeritus
Correct indeed