1. A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth verti
ID: 1730527 • Letter: 1
Question
1. A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall and reaches to a height of 2.10 m, as shown in the figure. The base of the ladder rests on a rough horizontal floor whose coefficient of static friction with the ladder is 0.800. An 80.0-kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall, as shown in the figure. a) What is the magnitude of the force that the floor exerts on the ladder? b) What is the magnitude of the friction force that the wall exerts on the ladder? c) What is the magnitude of the friction force that the floor exerts on the ladder?1. A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall and reaches to a height of 2.10 m, as shown in the figure. The base of the ladder rests on a rough horizontal floor whose coefficient of static friction with the ladder is 0.800. An 80.0-kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall, as shown in the figure. What is the magnitude of the force that the floor exerts on the ladder? What is the magnitude of the friction force that the wall exerts on the ladder? What is the magnitude of the friction force that the floor exerts on the ladder? a) b) c) 2.10
Explanation / Answer
b] Friction force by wall = 0 as it is smooth.
c]Distance of lower end from wall d = sqrt(2.5^2-2.1^2) = 1.356 m
Let normal by wall be N1 and N2 by ground,
balancing torque about lower end,
mg d/2 + Mgd = N1*2.1
N1 = [1.356*10*9.8/2 + 1.356*80*9.8]/2.1
= 538 N
N1 will be balanced by friction by ground,
Friction by floor = 538N
a] magnitude of force exerrted by floor = sqrt(538^2 + ((80+10)*9.8)^2) = 1033 N
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