Question #f3746

1 Answer
Jul 11, 2017

Answer:

See below.

Explanation:

Here

#L = # ladder length
#P = # ladder weight
#F_a = # applied horizontal force
#mu = # static friction coefficient
#H = # normal force in the vertical wall
#V = # normal force in the floor
#y_0 = # height ladder contact point
#x_0 = #width ladder contact point
#d = # applied force distance over the ladder
#sin(theta) = y_0/sqrt(y_0^2+x_0^2)#

Now we will develop the modeling for two cases:

1) #F_a# pushing to the left

#{(sum_X->H - F_a + mu V = 0),(sum_Y-> V - P - mu H = 0),(M_O-> x_0 V + F_a d sin(theta) - P (x_0/2) - y_0 H = 0):}#

2) #F_a# pushing to the right

#{(sum_X->H + F_a - mu V = 0),(sum_Y-> V - P + mu H = 0),(M_O-> x_0 V - F_a d sin(theta) - P (x_0/2) - y_0 H = 0):}#

Solving both cases we have for each case:

1) #F_a = 1/2(P ((mu^2-1) x_0 - 2 mu y_0) sqrt[x_0^2 + y_0^2])/( (mu x_0 - y_0) sqrt[x_0^2 + y_0^2]+d (1 + mu^2) y_0)#

putting figures we obtain the condition #F_a > 416.667# to the left

2) #F_a = 1/2(P ((mu^2-1) x_0 + 2 mu y_0) sqrt[ x_0^2 + y_0^2])/( (mu x_0 + y_0) sqrt[x_0^2 + y_0^2]- d (1 + mu^2) y_0)#

putting figures we obtain the condition #F_a > 38.889# to the right.