# Torque with a cylinder problem

• Tigers01
In summary, the problem involves a cylinder with weight W, where the coefficient of static friction for all surfaces is ⅓. The applied force P is equal to 2W. The question is to find the distance d for which counterclockwise motion is initiated by P. In order to solve this problem, it is necessary to consider the forces acting on the cylinder in both the horizontal and vertical directions. The vertical forces include P and W, while the horizontal forces include friction and the normal force. By setting up equations for each direction, the distance d can be calculated.

## Homework Statement

The cylinder of weight W is shown in the following diagram. The coefficient of static friction for all surfaces is ⅓. The applied force P = 2W.

I am working on number 11. Find the distance d for which counterclockwise motion is initiated by P. T = F r

## The Attempt at a Solution

I got d = ½ r but that is not an answer choice. Can someone else try this and tell me what you got?

#### Attachments

Tigers01 said:
I got d = ½ r

In your diagram, they both look like frictional forces and act at right angles to each other, but your first equation sums them.

haruspex said:
In your diagram, they both look like frictional forces and act at right angles to each other, but your first equation sums them.
f1 is the frictional torque that acts along the vertical and f2 is the frictional torque that acts on the horizontal.

The fictional force is equal to the normal times the coefficient and the normal is equal to all the forces acting down, so the normal is equal to W + 2W = 3W and then when I multiply by the coefficient of ⅓ I get Frictional force is equal to W. Then the f1 is W * 2r because the frictional torque acts along the edge.

Where the x is on the circle is where I used to find the torque of all the components.

Tigers01 said:
f1 is the frictional torque that acts along the vertical

Tigers01 said:
the normal is equal to all the forces acting down, so the normal is equal to W + 2W = 3W
Do you not see an inconsistency between those two statements?

haruspex said:
Do you not see an inconsistency between those two statements?
So what would the normal be for the horizontal?

Tigers01 said:
So what would the normal be for the horizontal?
Assign variables to the normal forces, N1, N2 say.
List all the vertical forces and write an equation relating them, then do the same for the horizontal forces.
At present you seem to be confusing the two directions.

haruspex said:
Assign variables to the normal forces, N1, N2 say.
List all the vertical forces and write an equation relating them, then do the same for the horizontal forces.
At present you seem to be confusing the two directions.

Like this?

Horizontal Friction F2:

F2 friction = Normal * ⅓ The normal for the horizontal is W + 2W = 3W
So F2 friction = 3W * ⅓ = W

Vertical Friction F1:

F1 friction = Normal * ⅓ The normal in on the vertical surface is from the fictional surface on the horizontal so the normal is W
Therefore, F1 = W * ⅓ = W/3

Tigers01 said:
The normal for the horizontal is W + 2W = 3W
Why? W and P are vertical forces.
Please do as I asked and make two lists of forces first, one for the vertical direction and one for horizontal.

haruspex said:
Why? W and P are vertical forces.
Please do as I asked and make two lists of forces first, one for the vertical direction and one for horizontal.

Horizontal Forces:
Just friction on the bottom points left

Vertical:
P (2W) points down
W points down
Friction on the vertical wall points up

Tigers01 said:
Horizontal Forces:
Just friction on the bottom points left

Vertical:
P (2W) points down
W points down
Friction on the vertical wall points up
You have left out the two normal forces.

haruspex said:
You have left out the two normal forces.
Horizontal Forces:
Just friction on the bottom points left
*And the normal* points right

Vertical:
P (2W) points down
W points down
Friction on the vertical wall points up
The normal points up

Tigers01 said:
Horizontal Forces:
Just friction on the bottom points left
*And the normal* points right

Vertical:
P (2W) points down
W points down
Friction on the vertical wall points up
The normal points up
Ok, so now turn each list into an equation.
Be sure to use distinct names for the two normal forces and for the two frictional forces, and make it clear which is which.
I shall be offline for an hour or two.

haruspex said:
Ok, so now turn each list into an equation.
Be sure to use distinct names for the two normal forces and for the two frictional forces, and make it clear which is which.

Horizontal Forces:
Just friction on the bottom points left: F 1 = Fn 2 * ⅓
*And the normal* points right: Fn 1 = F 1

Vertical:
P (2W) points down: P = 2W
W points down: W
Friction on the vertical wall points up: F 2 = Fn 1 * ⅓
The normal points up: Fn 2 = W + 2W

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Tigers01 said:
Horizontal Forces:
Just friction on the bottom points left: F 1 = Fn 2 * ⅓
*And the normal* points right: Fn 1 = F 1

Vertical:
P (2W) points down: P = 2W
W points down: W
Friction on the vertical wall points up: F 2 = Fn 1 * ⅓
The normal points up: Fn 2 = W + 2W
So you gave the normal force from the wall subscript 1 but the friction force from the wall subscript 2 (and vice versa for the ground)? Kind of confusing but it works I guess.
Tigers01 said:
The normal points up: Fn 2 = W + 2W
The equation you’re looking for describes that there is no net vertical force, right? So why is the friction from the wall not included?

Nathanael said:
So you gave the normal force from the wall subscript 1 but the friction force from the wall subscript 2 (and vice versa for the ground)? Kind of confusing but it works I guess.

The equation you’re looking for describes that there is no net vertical force, right? So why is the friction from the wall not included?
So Fn 2 should equal W + 2W - f 2 ?

Tigers01 said:
So Fn 2 should equal W + 2W - f 2 ?
Yes. Continue