A 90N dog climbs up a non-uniform (lambda = Cx^(3/2)), 100N ladder. The top of the ladder rests on the top of a frictionless fence that is 3m high and forms a 30 degree angle with the ground, which is rough and has a coefficient of friction of 0.60. How far up the ladder can the dog go before the ladder begins to slip?
I found the ladder to be 6m and the center of mass to be at 4.29m. I'm confused as to how to incorporate the center of mass into the problem. Thanks in advance!
Assume the 100 N weight of the ladder is applied at the computed location of the center of mass.
The normal (vertical) contact force at the lower point of contact with the ground is 190 N, and there is a horizontal friction force applied there that equals 190*0.60 = 114 N. set the total moment about the upper point of wall contact equal to zero, and solve for the unknown location of the pig.
So for the equation for the total, would this be correct?
(100)(4.29)cos30 - (90)(x)cos30 + (114)(6)sin30 = 0
No, the 100 and 90 weights produce a torque in the same direction. Those terms should have the same sign.
I get
(100)(4.29)cos30 + (90)(x)cos30 - (114)(6)sin30 = 0
I did not understand the
lambda = Cx^(3/2) part of your problem. I assume that it has something to do with ladder weight distribution but I have not verified your answer of 4.29.
Yes, it does have to do with the weight distribution. I'm confident with the 4.29 so up to here is all I need. Thank you for your help!
To determine how far up the ladder the dog can go before it begins to slip, we need to consider the forces acting on the ladder.
First, let's resolve the weight of the ladder into its components. The weight of the ladder is 100N, and it can be resolved into two components: one parallel to the ladder (W||), and the other perpendicular to the ladder (W⊥).
W|| = 100N * cos(30°) = 86.602 N
W⊥ = 100N * sin(30°) = 50 N
The normal force exerted by the fence on the ladder is equal and opposite to the perpendicular component of the weight, which is 50N.
Now, let's consider the frictional force acting on the ladder. The coefficient of friction is given as 0.60, and the rough ground exerts a frictional force opposing the motion of the ladder, which is the dog climbing up. The frictional force is given by:
Ffriction = μ * N
where μ is the coefficient of friction and N is the normal force. Therefore:
Ffriction = 0.60 * 50 N = 30 N
Since the frictional force opposes the motion up the ladder, it acts downward.
Next, let's consider the force the dog exerts on the ladder. The dog pushes the ladder forward, attempting to make it slip. This force (Fdog) acts upward.
Now, for the ladder to remain in equilibrium (not slip), the net force along both vertical and horizontal directions need to be zero.
In the vertical direction:
Sum of forces along the y-axis = Fupward - Fdownward = Fdog - W⊥
In the horizontal direction:
Sum of forces along the x-axis = F towards right - F towards left = Ffriction
Since the ladder is not slipping, the frictional force is equal to the force applied by the dog:
Ffriction = Fdog
Now, let's calculate the force applied by the dog (Fdog). The ladder is not slipping, so the force must be equal to or less than the maximum static frictional force:
Ffriction = μ * N
Fdog = μ * N
Substituting the value of N:
Fdog = 0.60 * 50 N = 30 N
Therefore, the force applied by the dog is 30 N.
Now, let's determine the torque around the base of the ladder. The torque applied by the dog (τdog) causes the ladder to rotate counterclockwise, and the torque caused by the weight of the ladder (τweight) causes the ladder to rotate clockwise.
τdog = Fdog * d (where d is the distance of the dog from the base of the ladder)
τweight = W|| * h (where h is the height of the center of mass of the ladder from the base)
Since the ladder is not slipping, the net torque must be zero:
τdog = τweight
Therefore,
Fdog * d = W|| * h
Substituting the known values:
30 N * d = 86.602 N * 4.29 m
Now, solving for d:
d = (86.602 N * 4.29 m) / 30 N
d ≈ 12.3 m
Therefore, the dog can go up approximately 12.3 meters before the ladder begins to slip.