First slide

Mechanical equilibrium force cases

Question

A mass of 1 kg is attached to the middle of a rope, which is being pulled from both the opposite directions. Taking g - 10 m/s², what is the minimum pull required to completely straighten the rope ?

Moderate

A

5N
B

10 N
C

20 N
D

infinity , rope cannot be straightened.

Solution

If A, B and C be three vectors acting at a point and their resultant is zero, then
$\frac{A}{\sin α} = \frac{B}{\sin β} = \frac{C}{\sin γ}$
where $α, β and γ$ are shown in figure.
To straighten the rope, $λ$ should be 180°. Now $α = β = 90^{\circ}$
$∴ A = B = \frac{C}{\sin 180} = \frac{1 \times 10}{0} = \infty$

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