First slide

Mechanical equilibrium force cases

Question

A block of mass 5 kg is suspended by a massless rope of length 2 m from the ceiling. A force of 50 N is applied in the horizontal direction at the midpoint P of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is (Take g = 10 ${ms}^{- 2}$ )m

Moderate

A

$30^{0}$
B

$40^{0}$
C

$60^{0}$
D

$45^{0}$

Solution

Let $θ$ be the angle made by the rope with the vertical in equilibrium.
The free body diagram of 5 kg block is as shown in fie.(b)
In equilibrium
$T_{2} = 5 g =$ 5 x l0 = 50 N
The free body diagram of the point P is as shown in Fig. (c).
In equilibrium
$T_{1} \sin θ = 50 N$ --------(i)
$T_{1} \cos θ = T_{2} = 50 N$ ------(ii)
Dividing (i) by (ii), we get
$\tan θ = \frac{50}{50} = 1$
$θ = \tan^{- 1} (1) = 45^{0}$

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