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.
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!
4 answers
So for the equation for the total, would this be correct?
(100)(4.29)cos30 - (90)(x)cos30 + (114)(6)sin30 = 0
(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.
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!