A spherical ball of weight 50N is suspended vertically by a string 500mm long. Find the magnitude and direction of the least force which can hold the ball 100mm above the lowest point . Also find the tension in the string at that point.
Answers
Answer:
A spherical ball of weight 50 N is suspended vertically by a string 50 cm long. Find the magnitude and direction of the least force, which ...
Explanation:
Answer:
Therefore the tension in the string at that point exists at 40N and the minimum force stands at 30N and the direction of force, tanθ = 3/4 along North East.
Explanation:
Given,
A spherical ball of weight 50N existed suspended vertically by string 500mm long.
To find : the magnitude and direction of the least force, which can hold the ball 100mm above the lowest point is..
also the tension in the string at that point is..
solution : draw free body diagram,
sinθ = 3/5 and cosθ = 4/5
at equilibrium state,
Tsinθ = Fcosθ
⇒T(3/5) = F(4/5)
⇒3T = 4F ....(1)
again, Tcosθ + Fsinθ = mg
⇒T(4/5) + F(3/5) = 50N
⇒4T + 3F = 250 ...(2)
from equations (1) and (2),
F = 30N and T = 40 N
Therefore the tension in the string at that point exists at 40N and the minimum force stands at 30N and the direction of force, tanθ = 3/4 along North East.
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