Finding the Coefficient of Static Friction

[Snape1830]

1. A wooden block of mass 29.0 kg sits on a horizontal table. A wire of negligible mass is attached to the right side of the block and goes over a pulley (also of negligible mass and frictionless), where it is allowed to dangle vertically. When a mass of 15.5 kg is attached to the dangling wire, the block on the table just barely starts to slide. What is the coefficient of static friction between the block and the table?

mass of block=29 kg
second mass = 15.5 kg
g=9.8 m/s^2
mu_s=?

Do I know my acceleration?

I honestly have no where to start. Please help!

 

[Pythagorean]

start with a free body diagram of the block on the table, give us the forces acting on the block.

 

[NascentOxygen]

Do I know my acceleration?

Acceleration is zero, since the blocks barely start to move.

But you do know the force that is causing this 'almost' movement.

 

[Snape1830]

start with a free body diagram of the block on the table, give us the forces acting on the block.

I don't really know how to draw a free body diagram (on the computer), but I know Ff, FN, gravity, and applied force and/or tension. Not quite sure about the last one. I just have no idea what equations to use.

 

[Snape1830]

Acceleration is zero, since the blocks barely start to move.

But you do know the force that is causing this 'almost' movement.

Gravity?

 

[NascentOxygen]

Gravity acting on the dangling block gives rise to the force I have in mind.

Gravity is usually thought of as an acceleration, rather than a force, pe se, as it has units of m/sec^2.

Perhaps you meant to write "weight"?

 

[NascentOxygen]

The two blocks are tied together, to effectively form one body of their combined masses, so they will accelerate together as one. This body is acted on by a force that tends to move it, but is opposed by friction.

 

[Pythagorean]

The two blocks are tied together, to effectively form one body of their combined masses, so they will accelerate together as one. This body is acted on by a force that tends to move it, but is opposed by friction.

yes; since it's not moving, you essentially have that the force on the left equals the force on the right. (friction vs. gravity).

Do you know the equations for each?

 

[Snape1830]

yes; since it's not moving, you essentially have that the force on the left equals the force on the right. (friction vs. gravity).

Do you know the equations for each?

Well the Fg=m*g and Ff=mu(Fn)

 

[Pythagorean]

And what is Fn?

 

[Snape1830]

And what is Fn?

Fn=m*a
But do I need to subtract something from Fn? I tried to find Fg and then I took Fn and subtracted Fg from it: Fn-Fg=ma (which equals 0)

 

[Snape1830]

And I calculated Fg to be 151.9 N

 

[Pythagorean]

there are two masses...

 

[Snape1830]

there are two masses...

Oh right...so I would need to add 15.5+29 and then multiply
By g or a depending on the equation?

 

[Snape1830]

there are two masses...

The thing I'm really confused about is the net force. I know that with the net force you have to take all of the forces into account. I just don't get what I subtract from what. Fn-Fg=ma to get Fn?
Fn-Fapp=ma to get Fapp?
Fapp-Ff=ma to get Ff?
I don't know what order, if there is an order. Because I think I tried it in this order and I got that Fg, Fn, and Fapp were all equal to 436.1 N (in whichever direction).

 

[NascentOxygen]

Fn=m*a

In addition to writing a formula, you need to also explain (in words and numbers) what these variables are. No one can confirm whether you are on the right track, otherwise. (What is m? What is a?)

Draw a diagram, and mark on it all the forces. This problem is one of the easier type; easier in the sense that every force you need to consider is acting either perfectly vertically, or perfectly horizontally, so no trigonometry is involved.

 

[Snape1830]

In addition to writing a formula, you need to also explain (in words and numbers) what these variables are. No one can confirm whether you are on the right track, otherwise. (What is m? What is a?)

Draw a diagram, and mark on it all the forces. This problem is one of the easier type; easier in the sense that every force you need to consider is acting either perfectly vertically, or perfectly horizontally, so no trigonometry is involved.

I figured it out! I have no idea how I did, but it became so clear there were two forces due to gravity (or weights). Thank you so so so much for your help! I really appreciate it!