A particle has three horizontal forces acting on it, as shown in the diagram. Show that cos α = 14 - 13 cos β 15 and find an expression for sin α . Use cos 2 α + sin 2 α = 1 to get an equation in β . Hence, find α and β .

Given three forces are acting on a particle as shown in the figure. Since the particle is in equilibrium, the net force on it will be zero. Equating the horizontal forces we have, 15 cos α + 13 cos β = 14 ................................................................(1) Or cos α = 14 - cos β 15 .............................................................................(2) Comparing the vertical component, we have, 15 sin α + 13 sin β = 0 or sin α = 13 15 sin β .........................)3) Since , sin 2 α + cos 2 α = 1 or sin 2 α = 1 - cos 2 α = 1 - 14 - cosβ 15 2 On simplification, we get, sin α = 29 - cos 2 β + 28 cosβ 225 1 2 .............................................(4) By solving we get sin α = 0 . 8 o r α = 53 . 1 ° . By equation (3) we get, sinβ = 15 13 sin α = 15 × 0 . 8 13 = 0 . 923 or β = 67 . 4 ° .

A particle has three horizontal forces acting on it, as shown in the diagram. Show that cos α = 14 - 13 cos β 15 and find an expression for sin α . Use cos 2 α + sin 2 α = 1 to get an equation in β . Hence, find α and β .

Given three forces are acting on a particle as shown in the figure. Since the particle is in equilibrium, the net force on it will be zero. Equating the horizontal forces we have, 15 cos α + 13 cos β = 14 ................................................................(1) Or cos α = 14 - cos β 15 .............................................................................(2) Comparing the vertical component, we have, 15 sin α + 13 sin β = 0 or sin α = 13 15 sin β .........................)3) Since , sin 2 α + cos 2 α = 1 or sin 2 α = 1 - cos 2 α = 1 - 14 - cosβ 15 2 On simplification, we get, sin α = 29 - cos 2 β + 28 cosβ 225 1 2 .............................................(4) By solving we get sin α = 0 . 8 o r α = 53 . 1 ° . By equation (3) we get, sinβ = 15 13 sin α = 15 × 0 . 8 13 = 0 . 923 or β = 67 . 4 ° .

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A particle has three horizontal forces acting on it, as shown in the diagram. Show that $\cos α = \frac{14 - 13 \cos β}{15}$ and find an expression for $\sin α$ . Use $\cos^{2} α + \sin^{2} α = 1$ to get an equation in $β$ . Hence, find $α$ and $β$ .

Important Questions on Forces in Two Dimensions

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IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 1

Find the component of the following force in the direction of the dashed arrow. Although it might seem clear from the diagram, make sure you specify whether the component is in the given direction or in the opposite direction.

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AS and A Level

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Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 1

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 1

Find the components of the following forces in the direction of the dashed arrow. Although it might seem clear from the diagram, make sure you specify whether the component is in the given direction or in the opposite direction.

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 1

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 2

Find the component of the following force perpendicular to the direction of the dashed arrow. Make it clear whether the component is in the perpendicular direction clockwise or anticlockwise from the direction given.

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AS and A Level

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Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 2

Find the component of the force perpendicular to the direction of the dashed arrow. Make it clear whether the component is in the perpendicular direction clockwise or anticlockwise from the direction given.

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 2

Find the components of the given force perpendicular to the direction of the dashed arrow. Make it clear whether the component is in the perpendicular direction clockwise or anticlockwise from the direction given.

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MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3B>Q 2

Find the component of the following given force perpendicular to the direction of the dashed arrow. Make it clear whether the component is in the perpendicular direction clockwise or anticlockwise from the direction given.

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A particle has three horizontal forces acting on it, as shown in the diagram. Show that cosα=14-13cosβ15 and find an expression for sinα. Use cos2α+sin2α=1 to get an equation in β. Hence, find α and β.

Important Questions on Forces in Two Dimensions

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A particle has three horizontal forces acting on it, as shown in the diagram. Show that $\cos α = \frac{14 - 13 \cos β}{15}$ and find an expression for $\sin α$ . Use $\cos^{2} α + \sin^{2} α = 1$ to get an equation in $β$ . Hence, find $α$ and $β$ .