A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.23 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 1.1 m along the ladder from the ladder’s footWhat is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 1.1 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.23 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 1.1 m along the ladder from the ladder’s footWhat is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 1.1 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦

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

good job you smart son of a gun you are smarter that einstein. 7

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05