A uniform ladder is 10 m long and weighs 290 N. In the figure below, the ladder

Question

A uniform ladder is 10 m long and weighs 290 N. In the figure below, the ladder leans against a vertical, frictionless wall at height h = 8.0 m above the ground. A horizontal force is applied to the ladder at a distance 6.0 m from its base (measured along the ladder). (a) If F = 50 N, what is the force of the ground on the ladder, in unit-vector notation? Fground = N If F = 150 N, what is the force of the ground on the ladder, in unit-vector notation? Fground = N Suppose the coefficient of static friction between the ladder and the ground is 0.38; for what minimum value of F will the base of the ladder just start to move toward the wall? Fmin = N thanks

Explanation / Answer

(c)
Summing forces in the vertical directions to zero shows us that the normal force of the floor on the ladder must equal the weight of the ladder or 270 N

The maximum friction force available will be
Ff = uFn
Ff = 0.38(290)
Ff = 110.2 N

if Rx is the horizontal reaction at the top of the ladder
we can sum horizontal forces to zero.

0 = F - Ff - Rx
Rx = F - Ff
Rx = F - 110.2

Recognizing that the geometry is a 3-4-5 triangle
we can sum moments about the ladder base to zero
let a clockwise moment be positive
Note: the figure given has a Force applied at the mid point of the ladder (5 m) while the question here has it applied at 6 m.

0 = F(6(4/5)) + 290(10(3/5) - Rx(8) = 0

substitute for Rx found earlier

0 = F(6(4/5)) + 290(10(3/5) - (8)(F - 110.2)
0 = F(24/5)) + 1740 - 8F + 881.6
F(8 - (24/5)) = 2621.6
F = 819.25 N
Fmin = 819 N

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