You lean 15­kg, 8.0­m ladder against a frictionless wall so it makes an angle of

Question

You lean 15­kg, 8.0­m ladder against a frictionless wall so it makes an angle of 70 degrees with the floor. The coefficients of friction between the ladder and the floor are ðs=0.40 and ðk=0.25. Just after you let go, the ladder stays in its position. What is the force of friction at this time?

I don't understand why 27N is the answer and not 59N. Please explain! Thanks in advance! :)

7. You lean 15-kg, 8.0-m ladder against a frictionless wall so it makes an angle of 70 degrees with the floor. The coefficients of friction between the ladder and the floor areu-0.40 and -0.25. Just after you let go, the ladder stays in its position. What is the force of friction at this time? a. 200 N 27 N C. 59 N d. 20 N e. ON

Explanation / Answer

Here ,

let the normal force from the wall is Nw

frictional force on the ladder is f

theta = 70 degree

mass , m = 15 Kg

length of ladder , L = 8 m

Balancing the moment of forces on ladder about the ground support point

Nw * L * sin(70) - m * g * (L/2) * cos(70) = 0

Nw * sin(70) - 15 * 9.8 * 0.5 * cos(70) = 0

Nw = 26.75 = 27 N

Now , balancng the forces in horizontal direction

Nw - f = 0

f = 27 N

hence , the force of friction at this time is 27 N .

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59 N is the maximum static friction , it is not necessary that maximum static friction act on an object all the time .static friction adjust itself so that object does not move.

hence , it is not 59 N

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