Please show FBD for this problem The 180lb man in the bosuns

The 180-lb man in the bosun\'s chair exerts a pull of 50 lb on the rope for a short interval. Find his acceleration. Neglect the mass of the chair, rope, and pulleys.

Solution

From Newton\'s second law:

1- Total forces acting on man:

Forces = m*a

T_rope + N - w_m = m_m*a,

T_rope + N = w_m + m_m*a

where: T_rope is the tension in the rope, N is the normal reaction of the chair on man, w_m is the man\'s weight & m_mis his mass.

2- Total forces acting on chair:

Forces = m*a

T_rope - N - w_c = m_c*a,

neglecting the chair\'s mass (& weight)

T_rope - N = 0

T_rope = N

where: T_rope is the tension in the rope, N is the normal reaction of the man on chair, w_c is the chair\'s weight & m_c is its mass.

By substituting:

T_rope + T_rope = w_m + m_m*a

2* T_rope = m_m*g + m_m*a =  m_m*(g+a)

Since the acceleration acts upwards, it gets a (-ve) sign. Due to the construction of the pulleys, only (1/8) of the force is needed to lift the chair.

2* T_rope =  (1/8)*m_m*(g-a)

Assume that the initial tension in the rope is sufficient to withstand the man\'s weight. The added tension in the rope is = the pull force/2.

2*(Pull/2) =  (1/8)*m_m*(-a)

Pull force = (1/8)*m_m*(-a)

a = (8*Pull force)/m_m = 1.89 ft/s² (acting upwards)     (answer)