A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-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 3.80 m along the ladder from the bottom. (horizontal force, vertical force)

(b) If the ladder is just on the verge of slipping when the firefighter is 9.10 m from the bottom, what is the coefficient of static friction between ladder and ground?

(c) What if? 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?

[tex]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[/tex]

Solving the equations gives:

[tex]F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}[/tex]

a)

[tex]F_R = 238N\\ F_N = 1310N[/tex]

b)

[tex]F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3[/tex]

c) Using the result from b and solving for [tex]r_F[/tex]

[tex]\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m[/tex]

ahirohit963 ahirohit963 · Answer 2 · 2021-10-23T15:27:36+02:00

a) N = 1310 N

b) [tex]\mu = 0.3[/tex]

c) [tex]\rm r_f = 2.4\;m[/tex]

Given :

length of ladder, [tex]\rm r_l[/tex] = 14 m

weight of ladder, [tex]\rm F_l[/tex] = 490 N

position of firefighter, [tex]\rm r_f[/tex] = 3.8 m

weight of firefighter, [tex]F_f[/tex] = 820 N

angle of ladder, [tex]\alpha = 63^\circ[/tex]

Solution :

Let force of the wall on the ladder be [tex]\rm F_w[/tex], force of friction on base of ladder be [tex]\rm F_r[/tex] and normal force on base of ladder be N.

Now, in x direction

[tex]\rm F_x = ma_x[/tex]

[tex]\rm F_w-F_r = 0[/tex]

In y direction,

[tex]\rm F_y = ma_y[/tex]

[tex]\rm N - F_f - F_l = 0[/tex]

[tex]\rm \tau_p= \overrightarrow{r} \times \overrightarrow{F}[/tex]

[tex]\rm \tau_p= r_f F_fcos\alpha + \dfrac{1}{2}r_lF_lcos\alpha - r_lF_wsin\alpha = 0[/tex]

[tex]\rm F_w = \dfrac{r_fF_f+0.5r_lF_l}{r_ltan\alpha }[/tex]

a) [tex]\rm F_r = 238 \;N[/tex]

[tex]\rm N = 1310 \;N[/tex]

b) [tex]\rm F_r = \mu N[/tex]

[tex]\mu = 0.3[/tex]

c) [tex]\mu =0.15[/tex]

[tex]\rm F_r = \mu N[/tex]

[tex]\rm r_f = 2.4\;m[/tex]

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