# A ladder with a length of 10.0 m and weight of 562.0 N rests against a frictionless wall, making an angle of 55.0◦ with the horizontal. ?

a) A ladder with a length of 10.0 m and weight of 562.0 N rests against a frictionless wall, making an angle of 55.0◦ with the horizontal. Find the horizontal force exerted on the base of the ladder by Earth when a ﬁreﬁghter weighing 819.0 N is 3.97 m from the bottom of the ladder. Answer in units of N.

b) Find the vertical force exerted on the base of the ladder by Earth. Answer in units of N.

c) If the ladder is just on the verge of slipping when the ﬁreﬁghter is 7.21 m up, what is the coeﬃcient of staticfriction between the ladder and the ground?

### 2 Answers

- Steve4PhysicsLv 71 month ago
If you change the figures, I think this is basically the same problem as the one solved here (second question down, numbered Ch 12.3 #13): https://www.cpp.edu/~skboddeker/131/131hw/ch12h.ht...

- FiremanLv 71 month ago
let the vertical & horizontal reaction force is By & Bx Newton respectively at the base of the ladder & The horizontal reaction force at the top is Tx Newton. Let the coefficient of friction at he base surface is µ, at the first condition, by balancing the vertical & horizontal forces:

=>By = 562 + 819 = 1381 N ----------------------Answer (b)

& Bx = Tx ---------------------------------(ii)

Now by taking the torque at the top:

=>562 x (10/2) x sin35* + 819 x (10-3.97) x sin35* + Bx x 10 x cos35*= By x 10 x sin35*

=>4444.40 + 8.19Bx = 5.74By ---------------------------(iii)

Now by taking the torque at the bottom:

=>562 x (10/2) x cos55* + 819 x (3.97) x cos55* = Tx x 10 x sin55*

=>3476.70 = 8.19Tx ---------------------------(iv)

=>Tx = Bx = 424.50 ------------Answer (a)

by (iii) & (iv) {as Tx = Bx}

=>4444.40 + 3476.70 = 5.74By

=>By = 1380 N -------------------------Answer(b)

Part2:

Now by taking the torque at the top:

=>562 x (10/2) x sin35* + 819 x (10-7.21) x sin35* + Bx x 10 x cos35*= By x 10 x sin35*

=>3392.14 + 8.19Bx = 5.74By ---------------------------(v)

Now by taking the torque at the bottom:

=>562 x (10/2) x cos55* + 819 x (7.21) x cos55* = Tx x 10 x sin55*

=>4998.71 = 8.19Tx ---------------------------(vi)

=>Tx = Bx = 610.34

by (v) & (vi) {as Tx = Bx}

=>3392.14 + 4998.71= 5.74By

=>By = 1461.82 N

But Bx = µ x By

=>µ = Bx/By = 610.34/1461.82 = 0.42 ------------------Answer (c)