A uniform stationary ladder of length l and mass m leans...

A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between the floor and ladder is Î¼. The ladder makes an angle Î¸ with respect to the floor. A painter of weight 1/2M stands on the ladder at a distance d from its base. a) Find an expression for the magnitude of the normal force N exerted by the floor on the ladder. b) Find an expression for the magnitude of the normal force Nw exerted by the wall on the ladder. c) Find the expression for the largest value of dmax for which the ladder does not slip. d) What is the largest value for d, in centimeters, such that the ladder will not slip? Assume that the ladder is 5.5m long, the coefficient of friction is 0.64, the ladder is at an angle of 44 degrees, and the ladder has a mass of 55kg.

Transcript

00:01 Hello everyone in this given view problem we have given a uniform ladder given the length l and mass and the coefficient of friction between the ladder will be given 0 .4 and the weight of the painter is 1 by 2 so we can write here let's hold the expression for magnitude of normal force and is equal to 3 by 2 but or we have given w is weight of painter it is given weight of painter so we can write we have given in the question 1 by 2 m length of the ladder is given l length of ladder l is given 5 .5 and coefficient of friction mu is given 0 .64 in mass of the ladder is given 55 kg in first part we have find the expression for the magnitude of normal force exerted so we can find here a we can write here magnitude of normal force is n so n minus w the normal force a this like this is ladder the normal force exerted in the upper direction and weight in the downward direction so n minus w the weight is w by 2 equals to 0 so n is equals to 3 by 2 and the weight of the ladder is given in the question w is given 1 by 2 so we can put here 3 by 2m so normal n is equal to 3 by 2 multiply m by 2 so normal will be 3 by 2 multiply mass of the letter is given 55 in the above divided by 2 so value of the normal is 41 .25 n and in the first part we filed the value of normal this is the answer of first part next, we can find expression for magnitude of the normal force exerted on the ladder, new force exerted on the ladder.

03:42 So normal force by wall on ladder.

03:58 So we can write here equals to friction force.

04:12 So normal force is equal to friction force.

04:16 Normal force on the wall by the ladder is equals to friction force so normal force on the wall equals a friction force is mu into n so n w is equal to the value of mu is 3 by 2 the value of n is fine w is given in the question 1 by 2m so we can write here 3 by 2 w is equal n value of n is fine so n w formal force on the wall is 0 .53 divided by 2 say 3 times 2 .5 3 multiply the value of n is 4 1 .25 so we can write here the value of n is 21 .86 newton this is the answer of b part.

05:34 Next in c we can find expression for larger value of d maximum which ladder does not slip...

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