If the external force F is applied at an angle $θ$ with the horizontal as shown in figure, the value of F for which friction force between the block of mass m and the wall is zero, is

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\frac{m g}{s i n θ}

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\frac{m g}{c o s θ}

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\frac{m g}{t a n θ}

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Solution

The correct option is A mg
Hence, option $(A)$ is correct answer.
Normal reaction = $F sin θ + m g cos θ$
Since the force is acting horizontally it is split into two component are along the plane and one perpendicular. Similarly we do for mg the weight.

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If the external force F is applied at an angle θ with the horizontal as shown in figure, the value of F for which friction force between the block of mass m and the wall is zero, is

The correct option is A mgHence, option (A) is correct answer.Normal reaction = Fsinθ+mgcosθSince the force is acting horizontally it is split into two component are along the plane and one perpendicular. Similarly we do for mg the weight.

If the external force F is applied at an angle $θ$ with the horizontal as shown in figure, the value of F for which friction force between the block of mass m and the wall is zero, is

The correct option is A mg
Hence, option $(A)$ is correct answer.
Normal reaction = $F sin θ + m g cos θ$
Since the force is acting horizontally it is split into two component are along the plane and one perpendicular. Similarly we do for mg the weight.