Recent content by Tigers01

So Fn 2 should equal W + 2W - f 2 ?

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

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

Horizontal Forces: Just friction on the bottom points left Vertical: P (2W) points down W points down Friction on the vertical wall points up

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...

So what would the normal be for the horizontal?

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...

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. Homework Equations T = F r...