(D) Let $T$ be the tension in the rope and $F$ be the horizontal force applied. The mass $m = 10\,kg$ is in equilibrium.Resolving the forces into hori

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(D) Let $T$ be the tension in the rope and $F$ be the horizontal force applied. The mass $m = 10\,kg$ is in equilibrium.Resolving the forces into horizontal and vertical components:Vertical component: $T\,\cos\,45^o = mg$Horizontal component: $T\,\sin\,45^o = F$Dividing the two equations: $\frac{T\,\sin\,45^o}{T\,\cos\,45^o} = \frac{F}{mg}$$\tan\,45^o = \frac{F}{mg}$Since $\tan\,45^o = 1$,we have $F = mg$.Given $m = 10\,kg$ and $g = 10\,ms^{-2}$,$F = 10 \times 10 = 100\,N$.

Class 11 Physics Newton's Laws of Motion and Friction The Common Forces and Equilibrium of Concurrent Forces

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$A$ mass of $10\,kg$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point,the rope deviates at an angle of $45^o$ from the vertical. If the suspended mass is at equilibrium,the magnitude of the force applied is .......... $N$ $(g = 10\,ms^{-2})$.

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Newton's Laws of Motion and Friction

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The Common Forces and Equilibrium of Concurrent Forces

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Write the condition for the translational equilibrium of a particle.

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For a particle to be in equilibrium,what must be the net force acting on it?

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Why are the states of a body at rest and a body moving with constant velocity considered equivalent in terms of net force?

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$A$ $100 \ N$ weight is suspended in a corner of a room by two cords $A$ and $B$ as shown in the figure. The tension in the horizontal cord $A$ is ............ $N$.

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$A$ mass $M \text{ kg}$ is suspended by a weightless string. The horizontal force required to hold the mass at $60^{\circ}$ with the vertical is

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Write the condition for the translational equilibrium of a particle.

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$A$ $100 \ N$ weight is suspended in a corner of a room by two cords $A$ and $B$ as shown in the figure. The tension in the horizontal cord $A$ is ............ $N$.

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