Answer:
a = 0.195 m/s²
T = 769.2 N
Explanation:
The sum of the forces (ΣF) acting on the man climbing up the rope is:
[tex] \sum F_{y} = m_{m}a_{m} [/tex]
[tex]T - m_{m}g = ma_{m}[/tex]
where [tex]m_{m}[/tex]: is the mass of the man, [tex]a_{m}[/tex]: is the acceleration of the man, T: is the tension and g: is the gravitational constant
[tex]T - (75 kg)(9.81 m/s^{2}) = (75 kg)a_{m}[/tex] (1)
Since the acceleration of the man is relative to the rope, the acceleration of the man is:
[tex] a_{r} = a_{m} - a [/tex]
[tex] a_{m} = a_{r} + a [/tex] (2)
where [tex]a_{r}[/tex]: is the acceleration of the man relative to the rope and a: is the acceleration of the rope = acceleration of the block A
By introducing equation (2) into (1) we have:
[tex]T - (75 kg)(9.81 m/s^{2}) = (75 kg)(0.25 m/s^{2} + a)[/tex] (3)
The sum of the forces acting on the block A is:
[tex] \sum F_{y} = ma [/tex]
[tex] m_{b}g - T = ma [/tex]
where [tex]m_{b}[/tex]: is the mass of the block A
[tex] (80 kg)(9.81 m/s^{2}) - T = (80 kg)a [/tex] (4)
Now, solving equations (3) and (4) for a and T, we have:
[tex] a = \frac{g(m_{b} - m_{m}) - (0.25)(m_{m})}{m_{b} + m_{m}} [/tex]
[tex] a = \frac{9.81 m/s^{2}( 80 kg- 75 kg) - (0.25 m/s^{2})(75 kg)}{80 kg + 75 kg} = 0.195 m/s^{2} [/tex]
[tex] T = m_{b}g - m_{b}a = (80 kg)(9.81 m/s^{2}) - (80 kg)(0.195 m/s^{2}) = 769.2 N [/tex]
I hope it helps you!
