A 140 m uniform ladder weighing 510 n rests against a...

A 14.0 m uniform ladder weighing 510 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal. (a) Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 820-N firefighter has climbed 3.70 m along the ladder from the bottom. horizontal force (b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between the ladder and ground? (c) If oil is spilled on the ground, causing the coefficient of static friction to drop to half the value found in part (b), what is the maximum distance (in m) the firefighter can climb along the ladder from the bottom before the ladder slips?

Transcript

00:01 In this question we are given that a 14 meter informed ladder being 510 newton rests against a frictionless wall and the ladder is making an angle of 65 degree with the horizontal alright this is the arrangement this is horizontal floor this is vertical wall right this is 14 meter and its weight is that is acting at the center it is 510 newton this is 65 degree angle and a person of 820 newton has climbed 3 .7 meter.

00:47 Alright, so this is the person that has climbed by 3 .7 meters.

00:55 And the weight of the person is 820.

01:00 We need to calculate the horizontal and vertical forces.

01:04 If this is the friction force f and this is the normal force and this is the normal force and this.

01:09 We need to calculate all these forces.

01:11 So first of all what we will do is we will try to balance the moment about, let us say this is 0 .0.

01:17 Sigma mo must be so 510 510 into 14 sorry not 14 it would be 7 it was up to the half length it would be 7 into 7 o 65 plus 820 into 3 .7 oh 65 minus n dash into 14 sine 65 that must be 0 so this and that's force would be equal to let us calculate 510 into 7 into post 65 plus 820 into 3 .7 into post 65 divided by 14 divided by sine 65 so this is coming out to be 2 1 or 220 this is the value of normal that way and as we are getting so if now, sigma f x must be equal to 0.

02:32 So if sigma fx is 0, then f must be equal to n -dus.

02:35 So that must be equal to 220.

02:37 And sigma fy must be equal to 0.

02:40 If the bowl was friction less, there was no friction on the bone, then n must be equal to the both the weights...

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