ladder physics problem

Question

An 80 kg person stands 0.75 m from the bottom of a ladder (i.e. d = 0.75 m). The ladder leans against a frictionless wall at a 60° angle. If the ladder is 50 kg and is 2.5 m long what is the friction due to the floor, in Newtons, on the ladder?

The ladder in the previous problem had a person standing on it. Assume this person climbs up so that he stands 2.0 m from the bottom of a ladder (i.e. d = 2.0 m). What is the force exerted by the wall, in Newtons, on the ladder?

Answer 1

Let:

the length of the ladder = L = 2.5 m

the mass of the ladder = m = 50 kg

the mass of the person = M = 80 kg

the distance the person stand from ladder bottom = d = .75 m

the angle ladder makes with the wall = θ = 60°

the normal force on the wall = N₂

the normal force on the floor = N₁

a) What is the friction, F due to the wall the floor on the ladder ?

From Newton's second law of motion, the x and y components of the forces involved are

x: ∑F_x = N₂ - F = 0 ==>N₂ = F = µN₁, where µ is the coefficient of static friction with the floor.

y: ∑Fy = N₁ -mg - Mg = 0 ==> N₁ = (m + M)g

Supposing the coefficient of friction is given, you can easily solve for the friction. but since we do not know it, we need to find it?

The net Torque, τ, on the system is zero, this means that

∑τ = 0 ==> mg(L/2)cosθ +Mgdcosθ - N₂Lsinθ = 0, solving for N₂ gives

N₂ = [(mgL/2 + Mgd)cosθ]/Lsinθ = (mg/2 +Mgd/L)/tanθ ---------------------------- A

Now the coefficient of friction, µ = N₂/N₁ = [(mg/2 +Mgd/L)/tanθ]/(m + M)g

plugging in the numbers and using g = 9.8m/s², you get

µ = 0.22, with this value you can find the friction, F = µN₁ = (.22)1274N = 280.3 N

b) The force exerted by the wall is N₂ = (mg/2 +Mgd/L)/tanθ, but now d = 2m

==> N₂ = 503.6 N

Answer 2

Hello Ana S.

You have a challenging problem here, because it requires the use of both torques and forces, as well as some simple trig functions.

The first step is to draw the ladder with nothing around it (no floors, no walls, no man) but acted upon only by forces. This is a free-body diagram. Since the ladder is not moving (we hope) we can say that it is in equilibrium, and so the sum of the forces in the x-direction = 0, in the y direction = 0, and the sum of the torques or moments (forces acting at a distance to spin objects around a given point) also = 0.

Second, you need to accurately draw the angle of the ladder, and calculate the sin and cosine of 60 deg.... this is easy. Now you need to set up all the equations of equilibrium, noting that since there is no friction from the wall, the force at the wall only acts perpendicular to the wall (all forces like this are called Normal Forces, and only act perpendicular to the surfaces which exert them). On the floor, however, there is a normal force AND a frictional force, and the normal force is perpendicular to the floor, while the frictional force acts solely parallel to the floor, and is the only force keeping the ladder from slipping and the man from spilling hot pink paint all over his mother's white shag carpet. (and from killing her tiny and useless white toy poodle Kiki, who is yipping endlessly from under the ladder, ignoring superstitions thereof...)

NOW you should be able to see (I hope you've already drawn the FBD! ... no need to draw Kiki (she is useless) or the paint bucket) that there are only two forces acting in the y-direction (vertically) so they must be equal and opposite) and there are only two forces acting horizontally, so same conclusion as above.

Lastly, you need to choose a point about which to take all your torques. This should be a point that eliminates one of the torques, so it can be at the point of the man, or the floor, or the wall. Also you need to choose a direction as a positive torque. Why not clockwise? (unless you are in the Southern Hemisphere, where toilets being flushed have the opposite direction of swirl!) Now every torque is made up of a force times a distance, but the only part of the force which acts is the part of it acting perpendicularly to the shortest line from the point chosen as "center". So now you need to use your trig functions (sin and cos of 60 deg) to find the part of the force acting as a torque around teh point chosen, and multiply it times the lever arm (perpendicular distance from the center) Once again, there are only two acting torques, so they must be ____ and ____, right? OK, if you can't get to the end from this so far, you need to call me.

Best wishes!

Michael W., Wyzant Pro Tutor