in 18.6 kg person climbs up a uniform ladder with negligible mass. the upper end of the ladder rest on a frictionless wall. the bottom of the ladder rest on a floor with a rough surface where the coefficient of static friction is 0.63. the angle between the horizontal and the ladder is theta. the person wants to climb up the ladder a distance of 1.6m along the ladder from the ladders foot. what is the minimum angle so that the person can reach the distance of 1.6m without having the ladder slip?

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aristocles aristocles · Answer 1 · 2019-02-15T01:48:53+01:00

let the normal force due to vertical wall is N1 and due to horizontal floor is N2

now by force balance we can say

[tex]N2 = mg[/tex]

[tex]N_2 = 18.6(9.8) = 182.3 N[/tex]

now friction force on the floor is given as

[tex]F_f = \mu N[/tex]

[tex]F_f = 0.63(182.3) = 114.8 N[/tex]

now by torque balance we will have

torque due to normal of vertical wall = torque due to weight of man

so here we have

[tex]N_1 (Lsin\theta) = mg(Lcos\theta)[/tex]

[tex]N_1 tan\theta = mg[/tex]

also we know that

[tex]N_1 = F_f = 114.8 N[/tex]

[tex]114.8 tan\theta = 18.6(9.8)[/tex]

[tex]\theta = 57.8 degree[/tex]

so above is the angle with the floor

ayfat23 ayfat23 · Answer 2 · 2021-11-29T15:16:29+01:00

The minimum angle required for the person to get to as distance of 1.6 m is

57.9°

The normal force can be separated into vertical and horizontal. Then to calculate the normal force we use

[tex]Normal force= (m*g)= (18.6*9.8)=182.3N[/tex]

F= (∪ [tex]N)[/tex]

[tex]=(0.63*182.3)=114.8N[/tex]

(F*tan θ [tex]= mg[/tex])

F= floor frictional force= 114.7N

m= 18.6 kg

[114.7*tan θ) [tex]=(18.6*9.81)][/tex]

114.7*tan θ =[tex]\frac{182.5}{114.7}[/tex]

[tex]=57.9[/tex]°

Therefore, the minimum angle is =57.9°