Find the reading of the spring balance shown in figure (5−E6). The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth.
Answers
Answer:
∴ The reading of spring balance will be 4.4 kg
Explanation:
Let A and B be the left and right blocks respectively.
let the acceleration of the 3 kg mass relative to the elevator be ‘a‘ in the downward direction.
From the FBD,
mAa=T-mAg-mAg/10 …1
mBa=mBg+mBg/10-T …2
Adding both the equations, we get:
a(mA+mB)=(mB-mA)g+(mB-mA)g/10
Substitute the value of masses in the equation we get:
9a=33g/10
⇒a/g=11/30 …3
Now, using equation (1), we get:
T=mA(a+g+g/10)
The reading of the spring balance = 2T/g
=(2/g) mA(a+g+g/10 )
⇒2×1.5(a/g+1+1/10)
=3(11/30+1+1/10)
= 4.4 kg
∴ The reading of spring balance will be 4.4 kg
Answer:
Answer:
∴ The reading of spring balance will be 4.4 kg
Explanation:
Let A and B be the left and right blocks respectively.
let the acceleration of the 3 kg mass relative to the elevator be ‘a‘ in the downward direction.
From the FBD,
mAa=T-mAg-mAg/10 …1
mBa=mBg+mBg/10-T …2
Adding both the equations, we get:
a(mA+mB)=(mB-mA)g+(mB-mA)g/10
Substitute the value of masses in the equation we get:
9a=33g/10
⇒a/g=11/30 …3
Now, using equation (1), we get:
T=mA(a+g+g/10)
The reading of the spring balance =
2T/g
=(2/g) mA(a+g+g/10
)
⇒2×1.5(a/g+1+1/10)
=3(11/30+1+1/10)
= 4.4 kg
∴ The reading of spring balance will be 4.4 kg