Torque with 2D center of mass

In summary, the problem involves a cube of mass M tilted against a wall, with no friction between the wall and the cube but sufficient friction between the cube and the floor to prevent slipping. The minimum coefficient of friction, μmin, is sought as a function of the angle θ when 0<θ<45°. The cube has two points of contact with the wall, and three forces are acting on the point of contact between the cube and the ground. These forces include N2, the normal force, μN2, the force parallel to the ground opposing any motion of the cube, and F2, the force parallel to the ground away from the wall. The force balance equations and torque equation are used to solve for the
  • #1
WilliamDaFoe
2
0

Homework Statement


A cube of mass M rests tilted against the wall as shown (see below). There is no friction between the wall and the cube, but the friction between the cube and the floor is just sufficient to keep the cube from slipping. When ##0\lt\theta\lt 45^\circ## find the minimum coefficient of friction ##\mu_{min}## as a function of ##\theta##

Hevoxpp.png


Homework Equations


[/B]
So, we have two points of contant with the wall. The point of contant between the vertical wall and the cube results in an unknown force ##N_1##. You then have three forces on the point of contant between the cube and the 'ground'. ##N_2## is the standard normal force. ##\mu N_2## would be a force parallel to the ground going towards the inner wall to oppose any action to move the cube and ##F_2## is the force parallel to the ground going away from the wall. From here force resolution gets us:

$$\sum F_x = N_2 -\mu N_1 = 0$$
$$\sum F_y = N_1 - mg(?)$$
$$\sum M_b = ??$$

The Attempt at a Solution


[/B]
So for the ##F_y## force my thinking is that because all of the weight of the cube will be pressing down at that one point we can get the $mg$ term. If not it will need to have a value of ##\frac{mgl}{\sqrt{2}}cos(\theta+45)##.

For the moment term I made an assumption that we could consider this box 1d and have the mass along the line so we get ##M_b=0=\frac{mgl}{2}cos(\theta)-lN_1 sin(\theta)##. This did not work.

I am thinking I need the CM to be at the center of the cube, in which case I have the mg term as mentioned but I am not sure how to find the ##M_b## term in this case.

additional stuff

This problem comes from the problem set for feynman and in the lecture for this he mentions classical field theory. I was also wondering if this problem could be solved using CFT and if it would be a worthwhile exercise to do so. Also, is it possible to solve this using virutal work? And how would I go about doing that? Looking for as many ways to solve this as possible to make sure I can do this later.
 
Last edited:
Physics news on Phys.org
  • #2
WilliamDaFoe said:
and ##F_2## is the force parallel to the ground going away from the wall.
That's a force the cube exerts on the ground. In analysing the cube, only consider forces acting on the cube.

For the LaTeX, either use doubled $ signs (which will put the equation on a line by itself) or doubled # signs.
 
  • #3
haruspex said:
That's a force the cube exerts on the ground. In analysing the cube, only consider forces acting on the cube.

For the LaTeX, either use doubled $ signs (which will put the equation on a line by itself) or doubled # signs.

Thank you, fixed.
 
  • #4
Ok, you altered the equation, but you still have F2 in the text, so it reads a bit strangely.

In your force balance equations, you seem to have crossed over N1 and N2 compared with how you defined them.

For your torque equation, please define the axis.
 

What is torque with 2D center of mass?

Torque with 2D center of mass is a concept in physics that describes the rotational force applied to an object around a point or axis. It takes into account the object's mass, distance from the pivot point, and the direction of the force.

How is torque with 2D center of mass calculated?

To calculate torque with 2D center of mass, you need to multiply the force applied to the object by the perpendicular distance from the pivot point to the line of action of the force. The resulting value is measured in Newton-meters (Nm).

What is the difference between torque with 2D center of mass and torque with 3D center of mass?

The main difference between torque with 2D center of mass and torque with 3D center of mass is the number of dimensions involved. Torque with 2D center of mass only takes into account the rotation around a single axis, while torque with 3D center of mass considers the rotation around all three axes.

What is the significance of torque with 2D center of mass in real-life applications?

Torque with 2D center of mass is important in many real-life applications, such as engineering, mechanics, and sports. It helps in understanding the stability, balance, and movement of objects, and is essential in designing structures and machines that can withstand rotational forces.

How does the center of mass affect torque in an object?

The position of the center of mass of an object determines the amount of torque it experiences. The farther the center of mass is from the pivot point, the greater the torque will be. This is because a greater distance results in a larger lever arm, which increases the rotational force.

Similar threads

  • Introductory Physics Homework Help
Replies
20
Views
324
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top