# Tag Info

0

Look at a free body diagram. With red and the contact normal forces, with pink the friction forces, and with gray the gravity forces. If in the end any of the friction forces come up to being negative, then flip the orientation.

1

The definitions are equal: Sum of external forces zero, sum of external torques is zero. This comes from classical mechanics. For a perfect ideal fluid, the external force density is the pressure gradient: $\mathbf f = -\nabla p$, and therefore, uniform pressure in a fluid means no external force on it, and then it is in mechanical equilibrium. So, its more ...

4

Okay, the system is in equilibrium so all the forces must be balanced. First consider the weights We have $W$ downwards from the centre of each of the 3 balls (assuming their centres of mass are at their geometric centres). We also have $N_A = N_B = \frac{3}{2}W$ upwards from the points of contact between the plane and balls $A$ and $B$ respectively. ...

0

Be careful - you are applying the reasoning about a static situation to a situation that is not static. This rod will be rotationally and linearly accelerating, so you can no longer assume the net force or net torque is zero - $\Sigma F_y = m a_y$, and $a_y \ne 0$. If the rod is instantaneously horizontal and at rest, it's not too hard to find $F_n$ at ...

2

You idea is exactly right. With that diagram, you need to figure out where the center of gravity is. The position of the center of gravity is a direct function of the number of bricks. In general, when you have $n$ bricks with equal offsets, the center of gravity is exactly midway between bricks $1$ and $n$ (by symmetry). The first brick is in contact with ...

1

Take for example, the case when the charge Q at the centre is negative, so the force is attractive. If it gets displaced even by a small value $\delta r$ towards either, the separation with this particular vertex would be $(\frac{l}{\sqrt 2}-\delta r)$, while it is slightly larger for the others. Due to the inverse dependence of the force on this ...

0

Consider the square to be in the x-y plane. Then see what happens when you displace the central charge in the z-direction. Of course, the answer will depend upon whether Q is positive or negative and possibly also on the direction of displacement.

Top 50 recent answers are included