Equilibrium and Moment of Inertia Question

[TBBTs]

Homework Statement

See attachment or here http://i.imgur.com/zBDa0.png

Homework Equations

Torque = FR

Moment of Inertia of Rod at End = I = 1/3 ML ^2

The Attempt at a Solution

http://i.imgur.com/3Avqh.png

 

[Doc Al]

For question A:
- Since the wall is frictionless, the force it exerts must be normal to the wall.
- Don't forget the normal force from the floor, when drawing your FBD.

For question B:
- Where's the center of mass?

 

[TBBTs]

For question A:
- Since the wall is frictionless, the force it exerts must be normal to the wall.
- Don't forget the normal force from the floor, when drawing your FBD.

For question B:
- Where's the center of mass?

Hi!
Thank you for replies.

A) So there are four forces?
Mg,Normal force from the wall and floor and Tension, but what about the direction?

B) Do I use Parallel Axis Theorm for this quesion?
Thanks!

 

[Doc Al]

Hi!
Thank you for replies.

Welcome to PF!

A) So there are four forces?
Mg,Normal force from the wall and floor and Tension,

Right.

but what about the direction?

What about it? Tell me the direction of each force.

B) Do I use Parallel Axis Theorm for this quesion?

If you like. (Not really needed.) But what's the moment of inertia of a point mass about an axis?

 

[TBBTs]

Welcome to PF!


Right.

What about it? Tell me the direction of each force.


If you like. (Not really needed.) But what's the moment of inertia of a point mass about an axis?

Hi!
For first part of my question, I use
1) sin30* N + Fh(Force of hinge) = Mg Fy
2) Fx: T=Ncos30
3) Torque: T*Lsin60 + Mg*Lcos(60) - NL =0

Are these equations right? I am so confused about how to determine the r of the torque (The torque formula is FR but how suppose I determine R?)

I am also confused about the direction of Torque,I know it supposed to use right hand rule and curl up my fingers to the direction its rotating,but since it is in equlibrium now,how suppose I know the way it is rotating?

Thanks!

 

[TBBTs]

Hi!
I tried Iend for two,so how about this?
(2/3)*(5+2)*0.5*0.5

Thanks!

 

[Doc Al]

For first part of my question, I use
1) sin30* N + Fh(Force of hinge) = Mg Fy

Please define N and Fh. (What hinge?)

Your diagram needs to be corrected. You have the normal force from the wall acting at some angle. It should be straight out from the wall, at 90 degrees.

2) Fx: T=Ncos30

Same problem.

3) Torque: T*Lsin60 + Mg*Lcos(60) - NL =0

Are these equations right? I am so confused about how to determine the r of the torque (The torque formula is FR but how suppose I determine R?)

When finding torque, R is the perpendicular distance from the line of action of the force to the pivot point.

I am also confused about the direction of Torque,I know it supposed to use right hand rule and curl up my fingers to the direction its rotating,but since it is in equlibrium now,how suppose I know the way it is rotating?

You can just make counterclockwise torques positive and clockwise torques negative. (The right hand rule is useful for problems in three dimensions. Here you don't really need it.)

Start by fixing that diagram and showing all forces acting on the ladder. (I count four forces.)