Question

Torque, I am trying to understand how they set up the torque in the following problem with a ladder.

Answer 1

Here is the problem and diagram.

Answer 2

I really need help just understanding my teachers solution. Thank YOU!!!

Answer 3

I'll use your teachers free body diagram as a basis.

We need to first define a system of kinetic equations that describe the forces on the ladder.

We have the center of gravity acting downwards at the middle of the length of the ladder. At the ground, we have the normal force acting upwards and the frictional force acting towards the wall. At the point where the ladder touches the wall, we have a normal force here. We are assuming no friction at the wall.

Let's assume the ladder is not moving. This means that newton's second law equates to zero. \[F = 0\]

We need to sum the forces in the x-direction, the y-direction, and about point O, where the ladder meets the ground. Let's say that positive torque is clockwise about this point.

\[\sum F_x = 0 = -F_w +F_f\]\[\sum F_y = 0 = - F_g + F_N\]\[\sum \tau_O =0 = -F_w l \sin(\theta) + F_g {l \over 2} \cos(\theta)\]
Let's take a closer look at this torque equation.

F_w doesn't act perpendicular to the ladder, but instead at an angle \(\theta\) off of perpendicular. Notice how the component of F_w that is perpendicular to the ladder is OPPOSITE the angle. Therefore, we use \(\sin(\theta)\) to find this component.

The center of gravity also doesn't act perpendicular to the ladder. Notice how the component of F_g that act perpendicular to the ladder is ADJACENT to the angle. Therefore, we use \(\cos(\theta)\) to find this component. It is also located half-way along the length of the ladder. Therefore, its moment arm is \(l/2\).

We can quickly observe that \(F_y\) reveals no new information. We already knew that F_N will equal the weight of the ladder. As this is by definition, assuming no other forces are present in the y-direction.

We can then use \(F_x\) and \(\tau_O\) to calculate the maximum angle.

This will be such that F_f = F_w.

Let's see how to set that up.

We know that F_f = F_w from the solution to \(F_y\).

We also know from our definition of static friction that\[F_f = \mu F_N = \mu mg\]Recall that \(F_y\) revealed that F_N = F_g.

Therefore, \[\mu mg l \sin(\theta) = mg {l \over 2} \cos(\theta)\]

\(mg\) cancels from both sides. Also, \(l\) will cancel. \[\mu \sin(\theta) = {\cos(\theta) \over 2}\]

\[\tan(\theta) = {\mu \over 2}\]