Question

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

Answer 1

Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

weight of ladder $F_L = 490N$

position of firefighter $r_F = 3.8m$

weight of firefighter $F_F = 820N$

angle of ladder $\alpha = 63$

Unknown:

force of the wall on the ladder $F_W$

force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$

Solving the equations gives:

$F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}$

a)

$F_R = 238N\\ F_N = 1310N$

b)

$F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3$

c) Using the result from b and solving for $r_F$

$\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m$