Question

The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 83.8 Non the wall to which it is attached. The tensions in the shorter cords are T2 = 69.5 N, T - 42.8 N, and T3 = 9.81 N. What are the masses of (a) disk A. (b) disk B. (c) disk C, and (d) disk D? (a) Number 1 Units (b) Number Units (c) Number Units (d) Number Units

Answer 1

T1 = 69.5 N

T2 = 42.8 N

T3 = 9.81 N

Now , the force acting on disk D are the tension T3 and the weight of the disk and the disk is in equilibrium which means that the T3 and weight will balance each other .

T3 = m_D * g (m_D= mass of disk D and g is the acceleration due to gravity)

9.81 = m_D * 9.81

m_D = 1 kg

Now , the forces acting on disk C are tension T2 , weight and T3 . T2 is acting upwards , weight and T3 are acting downwards .And the disk C is in equilibrium .Therefore ,

T2 = T3 + m_C * g (m_C = mass of disk C and g is the acceleration due to gravity = 9.8 m/s²)

42.8 = 9.81 + m_C * g

m_C * g = 42.8 - 9.81

m_C * g = 33.01

m_C = 33.01 / 9.8

m_C = 3.37 kg

Now , the forces acting on disk B are T1 ,T2 and the weight of the disk .The direction of T1 is upward and direction of weight and T2 is downwards .And the disk is in equilibrium .Therefore ,

T1 = T2 + m_B * g

69.5 = 42.8 + m_B *g (m_B = mass of disk B )

m_B*g = 69.5 - 42.8

m_B*g = 26.7

m_B= 26.7 /9.8

m_B= 2.73 kg

Now , the force acting on disk A are 83.8 N(tension due to the longer cord) ,T1 and weight of the disk . The tension due to the top cord is acting in the direction upward and T1 and weight are acting in the downward direction . And the disk A is in equilibrium . Therefore ,

83.8 = T1 + m_A *g (m_A= mass of disk A and g =9.8 m/s²)

83.8 = 69.5 + m_A*g

m_A*g = 83.8 - 69.5

m_A*g = 14.3

m_A= 14.3 / 9.8

m_A= 1.46 kg

m_A = 1.46 kg

m_B = 2.73 kg

m_C = 3.37 kg

m_D = 1 kg