Answer:
Q = arctan(7.1739) = 82.06
Step-by-step explanation:
Given:
- The mass of the person m = 20.3 kg
- The distance traveled up the ladder s = 1.1 m
- The gravitational constant g = 9.8 m/s^2
- The coefficient of static friction u_s = 0.23
- Total length of the ladder
Find:
The minimum angle θ, that would allow the person to climb without ladder slipping
Solution:
- Taking moments about point of ladder and wall contact A to be zero:
-F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0
- Taking Sum of vertical forces to be zero:
F_n,b - m*g = 0
F_n,b = m*g
- The frictional force F_f is given by:
F_f = u_s*F_n,b = u_s*m*g
- Plug the values back in:
- m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0
Simplify:
4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)
6.6*cos(Q) = 4*u_s*sin(Q)
tan(Q) = 6.6 / 4*u_s
- Plug in the values:
tan(Q) = 6.6 / 4*0.23
Q = arctan(7.1739) = 82.06
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