# Static and Kinetic Friction-NEED HELP!!

• Apr 5th 2009, 06:22 AM
victorfk06
Static and Kinetic Friction-NEED HELP!!
1.) What would be the acceleration of a 19 kg aluminum block if it were being pulled with a horizontal applied force of 300 N?

2.) In moving a 35..0kg desk from one side of a classroom to the other, a teacher finds that a horizontal force of 275 N is necessary to set the desk in motion and force of 195N is necessary to keep it in motion with constant velocity. What are the coefficients of (a) static friction and (b) kinetic friction between the desk and the floor?

3.) A 40kg crate is at rest on a level surface. If the coefficient of static friction between the crate and the floor is 0.69, what horizontal force is required to move the crate.

These problems I know how to do them but I get stuck...
• Apr 5th 2009, 06:54 AM
earboth
Quote:

Originally Posted by victorfk06
...

2.) In moving a 35..0kg desk from one side of a classroom to the other, a teacher finds that a horizontal force of 275 N is necessary to set the desk in motion and force of 195N is necessary to keep it in motion with constant velocity. What are the coefficients of (a) static friction and (b) kinetic friction between the desk and the floor?

3.) A 40kg crate is at rest on a level surface. If the coefficient of static friction between the crate and the floor is 0.69, what horizontal force is required to move the crate.

These problems I know how to do them but I get stuck...

Let $\displaystyle F_f$ denote the frictional force and N the force normal to the surface. Then

$\displaystyle F_f = \lambda \cdot N$

where $\displaystyle \lambda$ denotes the friction coefficient.

$\displaystyle F_f = \lambda \cdot N~\implies~\lambda = \dfrac{F_f}N$

to #2:

Plug in the values you know:

$\displaystyle \lambda_{stat} = \dfrac{275\ N}{35 \cdot 9.81\ N} \approx 0.8$

$\displaystyle \lambda_{kin} = \dfrac{195\ N}{35 \cdot 9.81\ N} \approx 0.568$

to #3:

The force normal to the surface is

$\displaystyle N = 40\ kg \cdot 9.81\ \frac{m}{s^2} = 392.4\ N$

Therefore the friction is:

$\displaystyle F_f = 0.69 \cdot 392.4\ N = 270.756\ N$
• Apr 5th 2009, 10:31 AM
victorfk06
Thanks now I understand but still working and stuck on the first one, it seems very easy but dont know what Im not getting....
• Apr 5th 2009, 10:57 AM
skeeter
Quote:

Originally Posted by victorfk06
1.) What would be the acceleration of a 19 kg aluminum block if it were being pulled with a horizontal applied force of 300 N?

$\displaystyle a = \frac{F_{net}}{m}$

were you given any info regarding other forces besides the applied force (like friction, etc.) ?