1. ## Mechanics question

A dog lies on the floor and refuses to move. When you pull the lead with a force
of 100 N and at an angle of 30 with the horizontal, the dog slides along the floor.
Determine the coefficient of friction, if the dog has a mass of 25 kg. (You can take

g = 10 m/s2.)

My working. [LaTeX ERROR: Convert failed]
Surely I need to have been told the acceleraion to work out F.

2. Originally Posted by poirot
A dog lies on the floor and refuses to move. When you pull the lead with a force
of 100 N and at an angle of 30with the horizontal, the dog slides along the floor.
Determine the coefficient of friction, if the dog has a mass of 25 kg. (You can take

g = 10 m/s2.)

My working. [LaTeX ERROR: Convert failed]
Surely I need to have been told the acceleraion to work out F.
$\mu_s = \dfrac{f_{s \, max}}{F_N}$

$\mu_s = \dfrac{100\cos(30^\circ)}{25g - 100\sin(30^\circ)}$

3. Originally Posted by poirot
My working. [LaTeX ERROR: Convert failed]

Forces are vectors so when you calculate the net force you are doing a vector sum. The simplest way to do this is to add the forces by components. In your work it looks like you just simply tried to add all the forces. This is not a correct approach.

I will use the usual xy coordinate system and say that the dog is being pulled up and to the right. In the FBD we have a weight w acting straight down (in the -y direction), a normal force $F_N$ acting directly upward, an applied force F acting up and to the right at a 30 degree angle, and a friction force f acting to the left (in the -x direction.)

$\sum F_x = F~cos(30) - f = ma$

Since we are looking for the coefficient of friction we are going to set a = 0 m/s^2. This gives us the maximum possible static friction, which is given by $f = \mu F_N$. So
$\sum F_x = F~cos(30) - f = 0$

In the y direction:
$\sum F_y = N + F~sin(30) - mg = 0$

Solve this equation for $F_N$ and solve the x equation for f and you can then apply skeeter's solution.

-Dan

4. Originally Posted by topsquark

Forces are vectors so when you calculate the net force you are doing a vector sum. The simplest way to do this is to add the forces by components. In your work it looks like you just simply tried to add all the forces. This is not a correct approach.

I will use the usual xy coordinate system and say that the dog is being pulled up and to the right. In the FBD we have a weight w acting straight down (in the -y direction), a normal force $F_N$ acting directly upward, an applied force F acting up and to the right at a 30 degree angle, and a friction force f acting to the left (in the -x direction.)

$\sum F_x = F~cos(30) - f = ma$

Since we are looking for the coefficient of friction we are going to set a = 0 m/s^2. This gives us the maximum possible static friction, which is given by $f = \mu F_N$. So
$\sum F_x = F~cos(30) - f = 0$

In the y direction:
$\sum F_y = N + F~sin(30) - mg = 0$

Solve this equation for $F_N$ and solve the x equation for f and you can then apply skeeter's solution.

-Dan
Thanks a lot for your help
I see my error, I was taking the friction to go up the slope so was resolving forces paralell to the slope. So frictional co-efficent = max friction/normal reaction? What does frictional co-efficent really mean ? I got sqrt(3)/4 by the way

5. Originally Posted by poirot
Thanks a lot for your help
I see my error, I was taking the friction to go up the slope so was resolving forces paralell to the slope. So frictional co-efficent = max friction/normal reaction? What does frictional co-efficent really mean ? I got sqrt(3)/4 by the way
Yes on both counts. The coefficient of (static in this case) friction is kind of an average of rough surfaces, "stickiness" of intermolecular forces, etc.) It's kind of messy when you get down to what causes it. But for (so far as I know of) all pairs of surfaces we can take the maximum static friction force to be some coefficient $\mu _s$ times the normal force. (Kinetic friction is always $\mu _k~F_N$.) The larger the coefficient of friction the harder it is to make the surfaces move across each other.

-Dan