ladder physics question

Question

In the figure above, a ladder of weight 200 N and length 10 m leans against a smooth wall (no friction on wall). A firefighter of weight 600 N climbs a distance x up the ladder. The coefficient of friction between the ladder and the floor is 0.5.

 

c. What is the maximum possible value of x if you don’t want the ladder to slip?

 

the angle between the floor and the ladder is 50 degrees

Answer 1

There are a lot of different things going on here, so lets sort them out a bit first and then tackle the question asked.

 

Let's call the base of the ladder point B. At this point, you can split the force between the ladder and floor into two parts: the normal force (F_NB) (vertically upward) between the two, and the frictional force (F_fr) (horizontal toward the wall) that keeps the ladder from slipping.

 

Let's call the top of the ladder where it rests against the wall point W. Because there's no friction, the only force here is the normal force (F_NW), horizontally outward from the wall.

 

Let's call the point where the firefighter stands point F. At this point, the weight of the firefighter (W_F) pushes down on the ladder with 600N of force vertically downward.

 

We have to assume the ladder is uniform, so that the center of the ladder (5m from the ground) is where the weight of the ladder (W_L) acts, vertically downward.

 

The whole system is at equilibrium, so the net forces in each direction, and net torque are all zero.

 

Using Newton's 2nd law for the vertical forces first:

F_NB - W_F - W_L = 0

F_NB - 600 - 200 = 0

F_NB = 800N

 

We can now calculate the frictional force:

F_fr = μ F_NB = 0.5 (800 N) = 400 N

 

Now we can balance the horizontal forces:

F_fr - F_NW = 0

F_NW = 400 N

 

Now for the tricky bit, balancing torques. We can choose any point to balance torques about, but since there are two torques at point B, that's a good choice, to make the equation simpler. Torque is the product of the component of each force perpendicular to the ladder multiplied by it's distance from the point of rotation.

 

If we let the distance the firefighter is up the ladder be x, then the torques and directions for each of the 5 forces relative to point B are:

τ_W = (10m) F_NW⊥ = (10m)(400N sin 50º) CCW (counterclockwise)

τ_L = (5m) W_L⊥ = (5m)(200N cos 50º) CW (clockwise)

τ_F = (x m) W_F⊥ = (x m)(600N cos 50) CW

τ_NB = (0m) F_NB⊥ = 0

τ_fr = (0m) F_fr⊥ = 0

(You should double check my sin/cos for each force. I'm not 100% confident they're correct relative to your figure, and might be reversed.)

 

We can balance these the same way we did the forces earlier. Let's call clockwise negative.

τ_W - τ_L - τ_F + τ_NB + τ_fr = 0

 

Now, finally,  you can solve for x!