Calculate Tension (Tp) and (TQ) (P is the mid point)
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you forgot to add figure . well,I added the figure.
between two blocks massive string is available.
mass of string as shown in figure is 0.2kg while point Q is placed massless string .
for tension at Q :
total mass is loaded by massless string , M = (2.9 kg + 0.2kg + 1.9kg) = 5kg
so, tension at Q , TQ = M(g + a)
[because system is accelerating upward with acceleration a = 0.2 m/s² so, psuedo force acts downward direction to form inertial frame ]
= 5 × (9.8 + 0.2) = 5 × 10 = 50N
for tension at P :
total mass is loaded on point P, M' = (1.9 + 0.1)kg = 2kg [because P is the middle point of massive string hence , mass of loaded string = half of mass of string= 0.1kg]
now, tension at P, TP = M'(g+a)
= 2 × (9.8 + 0.2) = 20N
between two blocks massive string is available.
mass of string as shown in figure is 0.2kg while point Q is placed massless string .
for tension at Q :
total mass is loaded by massless string , M = (2.9 kg + 0.2kg + 1.9kg) = 5kg
so, tension at Q , TQ = M(g + a)
[because system is accelerating upward with acceleration a = 0.2 m/s² so, psuedo force acts downward direction to form inertial frame ]
= 5 × (9.8 + 0.2) = 5 × 10 = 50N
for tension at P :
total mass is loaded on point P, M' = (1.9 + 0.1)kg = 2kg [because P is the middle point of massive string hence , mass of loaded string = half of mass of string= 0.1kg]
now, tension at P, TP = M'(g+a)
= 2 × (9.8 + 0.2) = 20N
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