Physics friction question

[jgalt42]

Hey guys,

So we have a wall and there is a rope attached to the wall on one end and a board on the other. The rope is attached to the board in a way that is 30 degrees to its horizontal, which makes the board prop up horizontally against the wall.

My question is, we have the tension (Tsin30) of the string pointing up and the gravitational force (mg) pointing down. Which direction is the vertical tension? Also what are the horizontal forces?

I was doing an EK question similar to this and it told me the the vertical frictional force was pointing down. Why?

Thanks!

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[matth87]

The Tension Force will have two components

Component 1 (Vertical):
Force Tension Pointing Upwards - (Tsin30)
Force Gravity Pointing Downwards - (mg)

Component 2 (Horizontal)
Force Tension Pointing Towards the Wall - (Tcos30)
Force Normal Pointing Away from the Wall - (should also be = to Tcos30)

If nothing is moving all the forces must balance out.

I dont see any frictional forces in this problem.

 

[jgalt42]

What if the board is touching the wall?

 

[IveyGirl]

it's touching the wall perpendicular to the wall, so there should be no component of friction acting in this scenario. friction only acts when two objects move against each other with a parallel component within the force (i.e. not perpendicular).

 

[matth87]

If the board is touching the wall like leaning against it and not bolted in then the component of friction would be static friction direct against intended motion.

Intended motion would be down due to gravity.

Maximum Force of Static Friction would point up wards, the same way as tension.

The net force would then be:
Force Gravity = (Tsin30) + Force friction
*** In this case friction will only be as big as needed to hold the plank in place.

If the dont give you a coefficen of friction, then you shouldn't expect to worry about friction in your problem.

 

[jgalt42]

Its EK 1001 Physics question 312

A 0.8m long sign extends from a wall supported by a rope as shown. If the sign has uniform density, what is the minimum coefficient of friction between the sign and the wall?

The rope is connected to the board at the 0.2m into the board. And its attached 60 degrees to the board on its horizontal.

 

[matth87]

The Torques should all cancel out.

Torque due to friction + Torque due to Tension in the Y = Torque due to the weight of the plank

r(Ff) + r(Ft)(sin60) = r(Fg)

the wall is .2 from the pivot point
the tension force verticle is .2 from point
the middle of the plank is .4m which is .2 from the pivot point

all the r's cancel

Ff + Ft(sin60) = Fg

Fg = weight of plank

Is that an answer?

 

[jgalt42]

Answer choices:

A. 0.17
B. 0.50
C. 0.87
D. 1.0

Answer is C

Keep in mind the rope is connected to the board 30 degrees vertical to the wall and 60 degree horizontal to the board.

Okay this is what the answer says

"This is an equilibrium problem. Remember, the frictional force f is smaller than or equal to (mu)N. We are looking for the smallest possible value of (mu). In order to avoid confusion, solve for the force f and then solve for (mu). If we choose the spot where the rope attaches to the sign, we can quickly see that, for the sign to be in rotational equilibrium, mg must equal f; the torque equation is 0.2mg = 0.2f. The forces upward equal the forces downward, so: Tcos30 = mg + f. Putting these two equations together, we have Tcos30 = 2mg >> T = 2mg/cos30. The forces left equal the forces to the right, so the normal force N equal Tsin30. The force of static friction is f <_ (mu)N = (mu) Tsin30. Earlier we found that Tcos30 = mg + f. Solving for f and substituting, we have: Tcos30 - mg <_ (mu) Tsin30 => (cos30/sin30) - (mg/Tsin30) <_ (mu). Earlier we found that T = 2mg/cos30. Substituting this value for T into the previous equation gives: (cos30/sin30) - (mgcos30/2mgsin30) <_ (mu) = > (cos30/sin30) - (cos30/2sin30) <_ (mu)"

My question is why is the gravitational force and the frictional force on the same side? (Tcos30 = mg + f). f = the frictional force