Question:
A body of weight \(W\) is hung with the help of a rope of negligible mass from a helicopter moving in a vertical plane. If the vertical upward acceleration and the horizontal acceleration of the helicopter are each equal to the acceleration due to gravity, then the tension in the rope is:
A body of weight \(W\) is hung with the help of a rope of negligible mass from a helicopter moving in a vertical plane. If the vertical upward acceleration and the horizontal acceleration of the helicopter are each equal to the acceleration due to gravity, then the tension in the rope is:
Show Hint
Whenever perpendicular accelerations are given, first find the resultant acceleration using Pythagoras theorem and then calculate the corresponding tension.
Updated On: Jun 12, 2026
- \(2W\)
- \(W\sqrt{2}\)
- \(W\sqrt{5}\)
- \(0\)
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The Correct Option is B
Solution and Explanation
Concept:
When a body is suspended from an accelerating support, the tension in the string depends on the resultant acceleration experienced by the body. If accelerations are present in mutually perpendicular directions, they must first be combined vectorially to obtain the effective acceleration.
The equation of motion is based on Newton's Second Law:
\[
\vec{T}+\vec{W}=m\vec{a}
\]
where \(T\) is tension, \(W=mg\) is the weight of the body and \(\vec a\) is the resultant acceleration.
Step 1: Identify the accelerations acting on the body. The helicopter has: \[ a_x=g \] (horizontal acceleration) and \[ a_y=g \] (vertical upward acceleration). Since these accelerations are perpendicular to each other, the resultant acceleration is \[ a=\sqrt{a_x^2+a_y^2} \] \[ a=\sqrt{g^2+g^2} \] \[ a=g\sqrt2 \]
Step 2: Determine the effective force required to produce this acceleration. The tension must balance the weight and simultaneously provide the horizontal acceleration. Horizontal component: \[ T_x=mg \] Vertical component: \[ T_y-W=mg \] \[ T_y=2mg \] However, it is simpler to work using vector addition. The effective acceleration relative to gravity becomes \[ \vec g_{\text{eff}} = g\hat{i}+g\hat{j} \] whose magnitude is \[ g_{\text{eff}} = g\sqrt2 \]
Step 3: Calculate tension. The string aligns itself along the resultant acceleration. Therefore, \[ T=m(g\sqrt2) \] Since \[ W=mg \] \[ T=W\sqrt2 \]
Step 4: Final answer. \[ \boxed{T=W\sqrt2} \]
Step 1: Identify the accelerations acting on the body. The helicopter has: \[ a_x=g \] (horizontal acceleration) and \[ a_y=g \] (vertical upward acceleration). Since these accelerations are perpendicular to each other, the resultant acceleration is \[ a=\sqrt{a_x^2+a_y^2} \] \[ a=\sqrt{g^2+g^2} \] \[ a=g\sqrt2 \]
Step 2: Determine the effective force required to produce this acceleration. The tension must balance the weight and simultaneously provide the horizontal acceleration. Horizontal component: \[ T_x=mg \] Vertical component: \[ T_y-W=mg \] \[ T_y=2mg \] However, it is simpler to work using vector addition. The effective acceleration relative to gravity becomes \[ \vec g_{\text{eff}} = g\hat{i}+g\hat{j} \] whose magnitude is \[ g_{\text{eff}} = g\sqrt2 \]
Step 3: Calculate tension. The string aligns itself along the resultant acceleration. Therefore, \[ T=m(g\sqrt2) \] Since \[ W=mg \] \[ T=W\sqrt2 \]
Step 4: Final answer. \[ \boxed{T=W\sqrt2} \]
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