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Standard XII

Physics

Vector Component

A force F is ...

Question

A force F is acting on a body and its component along x-axis is 60 N. The Force F is inclined at an angle $30^{\circ}$ as shown. Determine F.

F = 30 \sqrt{3} N

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F = 40 \sqrt{3} N

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F = 50 \sqrt{3} N

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F = 60 \sqrt{3} N

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Solution

The correct option is B $F = 40 \sqrt{3} N$
Horizontal component of force = $F cos θ = 60$
$F = \frac{60}{cos θ} = \frac{60}{cos 30^{\circ}} = 40 \sqrt{3} N$

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A force F is acting on a body and its component along x-axis is 60 N. The Force F is inclined at an angle 30∘ as shown. Determine F.

The correct option is B F=40√3 NHorizontal component of force = Fcosθ=60 F=60cosθ=60cos30∘=40√3 N

A force F is acting on a body and its component along x-axis is 60 N. The Force F is inclined at an angle $30^{\circ}$ as shown. Determine F.

The correct option is B $F = 40 \sqrt{3} N$
Horizontal component of force = $F cos θ = 60$
$F = \frac{60}{cos θ} = \frac{60}{cos 30^{\circ}} = 40 \sqrt{3} N$