A uniform rod XY of weight 2.0N has a length of 80 cm.
The rod is suspended by a thread 20 cm from end X. A weight of 5.0 N is suspended from end X.
A student hangs a 6.0 N weight on the rod so that it is in equilibrium.
What is the distance of the 6.0 N weight from end X?
Answers
Answer:
30 cm
Explanation:
Hope it will help you.
If you don't understand the solution, then do let ne know.
Answer:
The distance of the 6 N weight from end X is 30 cm.
Explanation:
Assumption,
Let the distance from the hanging point to the 6 N weight be 'a'
Given,
Weight of the rod(W) = 2 N
Length of the rod (L) = 80 cm
Weight of a body suspended at end X (Wₓ) = 5 N
distance from the suspended point to end X (Lₓ) = 20 cm
Weight of the body suspended at distance 'a' from the point of suspension (Wₐ) = 6 N
To find,
The distance from the hanging point to the 6 N weight be 'a'
Calculation,
Applying moment of torque at suspension.
T = r × F
T₁ = 5 N × 20 cm
T₂ = torque due to weight + torque due to 6 N weight.
T₂ = 2 N × (40 - 20) cm + 6 N × a
For equilibrium condition,
T₁ = T₂
5 N × 20 cm = 2 N × 20 cm + 6 N × a
100 Ncm - 40 Ncm = 6 N × a
60 Ncm = 6 N × a
a = 10 cm from the point of suspension.
From point 'X' the distance is (10 cm + 20 cm) = 30 cm.
Therefore, the distance of the 6 N weight from end X is 30 cm.
#SPJ2