A uniform ladder of mass m = 40 kg and length 1 = 10 m is leaned against a frictionless wall. A person with mass M = 80kg stands on the ladder at a distance x = 7 m from the bottom, as measured along the ladder. The ladder makes a 30° degree angle with the floor. The floor is not frictionless and the ladder does not move. What is the normal force that the wall exerts on the ladder? 10 m 7 m 30° Image size: S M L Max +
The answer is 1290N, but I am not entirely sure how they got there. However, I noticed that when I calculated the torques for the person and that of the ladder, my answer would be correct if I multiplied by sin(60) or cos (30) instead. Using the sum of these values, I would divide it by the length of the ladder and sin(30), which would give the correct answer, but I'm a bit confused as to why I need to take sin(60) or cos(30) to calculate the torques for the weight of the ladder and the human, but then use sin(30) when calculating the torque for the perpendicular force. I would appreciate a work through of the problem and some explanation for my difficulties above, thank you.
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