Recent content by ChessEnthusiast

I see. So, to recap: There are three forces acting on the wrench: gravity the force we apply the force the bolt applies The last force does not contribute to the net torque. There are three factors contributing to it: gravity the force we apply the friction torque Is it correct so far?

Let's say we want to change a wheel in a car. We want to remove bolts fastening the wheel using this tool: I have also drawn a diagram of the forces in operation: Now, from experience I can say that the point of rotation of the wrench will be the blue point. Now, trying to determine the...

Suppose that I know that the object will not slide - is it, then, sufficient to find the angle that causes the line of action to leave the base of the cuboid?

Let's say I have a cuboid with sides $a$, $a$ and $3a$. The distance of the line of action of the force of friction from the center of gravity is $\frac 3 2 a$ As for the reaction force, the maximal distance will be $\frac a 2$. We are considering the static case. Therefore, the force of...

We have a cube on an inclined plane. The tipping condition is the presence of an unbalanced torque relative to the center of mass (contributing forces are: the normal force and the force of friction). However, is this conditions equivalent to the previous one: The line of action of the force of...

So, the perpendicular distance from the center of mass to the bottom surface is $a/2$. Both, the normal force and friction act on the perpendicular arm of $a/2$. Letting clockwise be the positive direction, the torque of the normal force is $$\tau_N = \frac{a}{2}mg \cos(\theta)$$ Ant the...

So now, as I try to compute the net torque with respect to the center of mass, I end up with only one contributing force, the normal force. The force of friction is parallel to the ground and the force of gravity acts at a radius of 0, and so generates no torque. The torque generated by the...

I was thinking about the rightmost point in contact with the ground, but now I know it was not a good choice. No, the torque generated by the force of friction will equal zero, because it's parallel to the ground.

I'm not sure where the frictional force is applied. It seems to me that it should be the rightmost point of the cube in contact with the ramp, but I'm not sure about it.

So, since - in the critical position - the torque applied by the normal force is 0, there are still three forces that apply a nonzero torque 1. Vertical component of the force of gravity 2. Horizontal component of the force of gravity 3. Friction How do I determine the torque of the friction...

@Dale Intuition tells me that the normal force should be "shifted" to the right, but I am unable to determine the exact distance. Will the normal force, right before tipping, act directly on the point I chose as the axis?

Why doesn't the one I chose work?

@CWatters When $\mu = \tan(\theta)$ so that's around 26 degrees. I have thought about this problem a little bit more and I've come up with another approach. The cube will tip once it gets a non-zero torque in the clockwise direction with respect to the axis I marked. There are several...

What's the minimal angle of a ramp such that a cube with side 10 cm placed on it will tip over? The coefficient of friction between the cube and the ramp is 0.6. I have given it some thought and I have come to certain conclusions: I think that the cube will tip over once the force of gravity...

We have two identical balls sitting at the same height. One of them is released without applying any force (it falls freely) and the other one is given a non-zero horizontal velocity. Which of them will hit the ground first? The force of air resistance is proportional to the square of the...