Question

The man (75 Kg) is climbing up the rope with acceleration 0.25 m/s2 relative to the rope. MA = 80 Kg. What is the rope tension and aA? Ans: aA = 0.195 m/s2 down, T = 769 N

Answer 1

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:

$\sum F_{y} = m_{m}a_{m}$

$T - m_{m}g = ma_{m}$        

where $m_{m}$: is the mass of the man, $a_{m}$: is the acceleration of the man, T: is the tension and g: is the gravitational constant                                

$T - (75 kg)(9.81 m/s^{2}) = (75 kg)a_{m}$        (1)                                              

Since the acceleration of the man is relative to the rope, the acceleration of the man is:

$a_{r} = a_{m} - a$

$a_{m} = a_{r} + a$          (2)

where $a_{r}$: 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:

$T - (75 kg)(9.81 m/s^{2}) = (75 kg)(0.25 m/s^{2} + a)$   (3)

             

The sum of the forces acting on the block A is:

$\sum F_{y} = ma$    

$m_{b}g - T = ma$        

where $m_{b}$: is the mass of the block A

$(80 kg)(9.81 m/s^{2}) - T = (80 kg)a$            (4)    

Now, solving equations (3) and (4) for a and T, we have:

$a = \frac{g(m_{b} - m_{m}) - (0.25)(m_{m})}{m_{b} + m_{m}}$                                                                                    

$a = \frac{9.81 m/s^{2}( 80 kg- 75 kg) - (0.25 m/s^{2})(75 kg)}{80 kg + 75 kg} = 0.195 m/s^{2}$

$T = m_{b}g - m_{b}a = (80 kg)(9.81 m/s^{2}) - (80 kg)(0.195 m/s^{2}) = 769.2 N$

I hope it helps you!