Now that you have read the section of my page on Ladder Problems, here are some pre-solved problems for you to examine to further your understanding of the concepts in the ladder problem section. After you examine these, you'll be able to try some out for yourself.
A ladder that is 4.0 m long and has a mass of 5.0 kg is placed against a frictionless wall and forms an angle of 60 degrees with the horizontal as shown in the above diagram. A man with a mass of 60.0 kg climbs 3.1 m up the ladder. What frictional force is required to keep the ladder from slipping?
The ladder is in equilibrium at present so we know that the net force acting in the X and Y direction must be zero. Therefore we can deduce:
Fa + Ff = 0
Fa = Ff
And:
Fg(man and ladder) + Fn = 0
Fg = Fn
We know that the Fg of the ladder equals 49 N and the Fg of the man equals 588 N, so the total Fg force equals 637 N. From this we know that Fn must equal 637 N as the two are equal. We do not know Ff or Fn, but if we knew Fn, we would know the value of Ff as they must be equal.
To get some other values in this problem, we turn to the other statement of equilibrium, that the net torque must equal zero. Now, we pick a pivot point. I pick the bottom of the ladder since it will eliminate two of the forces I would have to deal with. Some might say, well the force of friction that we need to solve for is not going to be included in the torques. This is true, but Fa is and we know thanks to the laws of equilibrium, that Fa equals Ff. If we find Fa, we have Ff, so lets do it the easy way.
Our counterclockwise torques are generated by the man's weight and the ladder's weight. The one clockwise torque is Fa. Now all we need is the lever arm values which we can get using trig.
Lever Arm of Fa: Sin60 = x/4.0 m
x = 3.5 m
Lever Arm of Fgladder: Cos60 = x/2.0 m
x = 1.0 m
Lever Arm of Fgman: Cos60 = x/3.1 m
x = 1.55 m
Now we can use the torque equilibrium statement or equation to solve for Fa:
Torque CW = Torque CCW
(49 N)(1.0 m) + (588 N)(1.55 m) = (Fa)(3.5 m)
960.4 N*m = (Fa)(3.5 m)
270 N = Fa
Since Fa and Ff are equal then we know that:
Ff = 270 N
A 5.2 m long ladder with a mass of 2.0 kg is placed against a frictionless wall as shown in the above diagram. A man with a mass of 65 kg climbs 3.4 m up the ladder. If the coefficient of static friction is 0.34, what angle does the ladder form with the ground?
Again the ladder is an example of an equilibrium solution so all laws of equilibrium can be applied.
We know, thanks to the laws of equilibrium, that the net torque and forces equal zero. If we could get a torque equation, we would need to use the angle for the lever arm computations and thus we would be able to solve for the angle. What values do we know that could contribute to this objective?
We know that the Fgladder equals 19.6 N and that the Fgman equals 637 N, and therefore we know that thier sums must be equal to Fn as the ladder is in equilibrium, thus Fn is equal to 656.6 N. We also know the coefficient of static friction which we can use to find Ff through the following equation learned in a previous chapter:
Coefficient of Static Friction = Ff/Fn
0.34 = Ff/656.6 N
223 N = Ff
From this we also know that since Fa is equal to Ff, that Fa is equal to 223 N.
Now in order to employ torques we need to pick a pivot point and get some lever arm values. The pivot point again shall be at the bottom of the ladder and we'll calculate the lever arm values keeping that in mind:
Fg Ladder: CosA = x/2.6 m
(CosA)(2.6 m) = x
Fg Man: CosA = x/3.4 m
(CosA)(3.4 m) = x
Fa: SinA = x/5.2 m
(SinA)(5.2 m) = x
We now have values for all of the lever arms and so we can now solve for the value of the unknown angle A using the fact that the net torque is zero, we can employ the following equation:
Torque CW = Torque CCW
(CosA)(3.4 m)(637 N) + (CosA)(2.6 m)(19.6 N) = (SinA)(5.2 m)(223 N)
(CosA)(2165.8 N*m) + (CosA)(50.96 N*m) = (SinA)(1159.6 N*m)
(CosA)(2216.76) = (SinA)(1159.6)
2216.76 = (TanA)(1159.6)
1.91 = TanA
62 degrees = A
Now that you have examined some solved problems, head to the Ladder Problems Page and try some for yourself.
