Why an object with double length have the same friction? [closed]

I didn't really got the highlighted part.

• Well, might you explain what exactly you didn't get about the highlighted part? Also, please include quotes as actual text, not as pictures. Apr 30, 2016 at 11:18
• How can I convert into actual test directly from a pdf,Btw the question is almost answered by Snyder005 Apr 30, 2016 at 12:08
• @NihalJalaluddinP Do you have a keyboard handy? You could type it. Apr 30, 2016 at 17:34
• @Asher how to write as quotes am using this my iPad May 1, 2016 at 1:30
• Type a right angle bracket > and then the text you want to quote May 1, 2016 at 1:50

The force of friction is defined as $F_f = \mu N$, where $N$ is the normal force. In the case of a flat surface free of external forces, you can use Newton's laws to determine that $N = mg$, where $m$ is the mass of the object. Notice that we have made no reference to the objects size, or area of contact. This is because in these examples we have essentially approximated the object as a point particle, and the forces as being applied to the objects center of mass. Probably not enough to convince you however.
Here is an illustration. Let's suppose we have a block of mass $m$. We can treat the frictional force then as $F_f = \mu mg$. Now lets say we cut that block in half, horizontally and then set the halves next to each other. The frictional force on each half is now $F_{f,half} = \mu mg/2$ (since each half weighs half as much). Now pretend we've magically welded the two halves together, so as they are now the same object. The frictional force should be the sum of the force on the two halves (since they move together), and will be $2F_{f, half} = F_f$. As you can see, subdividing the object and spreading it out has no effect on the total friction, since we are also decreasing the mass of each subdivision, which decreases the normal force, which decreases the friction of the subdivision.