# Simple but unsure, compound questions

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)

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

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

Hootenanny
Staff Emeritus
Gold Member
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
What is the equation for the maximal static frictional force?

What is the equation for the maximal static frictional force?

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

Hootenanny
Staff Emeritus
Gold Member
mu = F/N right?
This is correct.
So therefore F=(mu)N -> F = (mu)mg?
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.

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.

omg, how did I miss that 0_0

thank you, everything is 100% clear now =P

Hootenanny
Staff Emeritus
A pleasure 