# Simple but unsure, compound questions

1. Sep 11, 2008

### vysis

A boy is pulling on a box at a 31 degree angle using 21N of force. The box remains stationary. The co-efficient of static friction is 0.55. What is the minimum mass?

When I use the component method to solve this problem, do I need to calculate the mass for both the Horizontal and the Vertical and add them up? Or just the Horizontal?

(I ask this because when I solved for horizontal, the textbook said it was wrong. However, when I solved for both vertical and horizontal, I got the right answer)

2. Sep 11, 2008

### Hootenanny

Staff Emeritus
I'm not quite sure what you mean, could you post your working?

3. Sep 11, 2008

### vysis

Basically:

givens
v(1) = 0m/s
F(t) = 21N (31 degree)
(mu kinetic) = 0.5
(mu static) = 0.55
a = 0m/s^2

X component
Fnet=ma
F(t) - F(f) = ma
F(t) = (mu)Mg
m = F(t) / (mu)g
m= 18N / 0.55*9.8
m = 3.3kg

Y Component

F(net) = ma
F(t) = Fg
F(t) = m/g
m = 21Sin31 / 9.8
m = 1.1kg

the answer from the textbook is 4.4kg

4. Sep 11, 2008

### Hootenanny

Staff Emeritus
Here's your error:
What is the equation for the maximal static frictional force?

5. Sep 11, 2008

### vysis

isn't the equation (mu)mg? mu = F/N right? So therefore F=(mu)N -> F = (mu)mg?

6. Sep 11, 2008

### Hootenanny

Staff Emeritus
This is correct.
But this isn't. The normal force isn't necessarily equal to the weight of the object. This is only case on a horizontal surface when the only two forces acting in the vertical plane is the object's weight and the normal force.

You should start by examining the forces acting in the vertical plane.

7. Sep 11, 2008

### vysis

omg, how did I miss that 0_0

thank you, everything is 100% clear now =P

8. Sep 11, 2008

### Hootenanny

Staff Emeritus
A pleasure

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