A wooden block is held in position by three horizontal forces, as shown in the diagram. One acts to the left with force 56 N . One acts with force F at an angle θ . where sin θ = 3 5 , , above the rightwards horizontal. One acts with force G at an angle φ , where sin φ = 15 17 , , below the rightwards horizontal. Find F and G .

Given Two forces F . G acting on a block making angles θ , φ with the horizontal as shown, to the right side A force of 56 N is acting in the opposite direction. Given that sin θ = 3 5 , sin φ = 15 17 so we have, cos θ = 4 5 and cos φ = 8 17 . At equilibrium, the net force on the block is zero. Equating the vertical forces, we get, F sin θ = G sin φ or F × 3 4 = G × 8 17 or F = 32 21 G .....................................(1) Equating the horizontal forces, we have, F cos θ + G cos φ = 56 N Substituting the value in this equation, we have, 32 21 G × 4 5 + G × 8 17 = 56 or 128 105 G + 8 17 G = 56 Or 1 . 22 G + 0 . 47 G = 56 or G 33 . 13 N ...............................(1) And F = 33 . 13 × 32 21 = 50 . 5 N .................................................(2)

A wooden block is held in position by three horizontal forces, as shown in the diagram. One acts to the left with force 56 N . One acts with force F at an angle θ . where sin θ = 3 5 , , above the rightwards horizontal. One acts with force G at an angle φ , where sin φ = 15 17 , , below the rightwards horizontal. Find F and G .

Given Two forces F . G acting on a block making angles θ , φ with the horizontal as shown, to the right side A force of 56 N is acting in the opposite direction. Given that sin θ = 3 5 , sin φ = 15 17 so we have, cos θ = 4 5 and cos φ = 8 17 . At equilibrium, the net force on the block is zero. Equating the vertical forces, we get, F sin θ = G sin φ or F × 3 4 = G × 8 17 or F = 32 21 G .....................................(1) Equating the horizontal forces, we have, F cos θ + G cos φ = 56 N Substituting the value in this equation, we have, 32 21 G × 4 5 + G × 8 17 = 56 or 128 105 G + 8 17 G = 56 Or 1 . 22 G + 0 . 47 G = 56 or G 33 . 13 N ...............................(1) And F = 33 . 13 × 32 21 = 50 . 5 N .................................................(2)

A block with weight 44 N is held in equilibrium by two ropes, one with tension, T 1 , acting at angle sin - 1 15 17 to the upwards vertical and the other with tension, T 2 , acting at angle sin - 1 5 13 to the upwards vertical. Find T 1 and T 2 .

Given the mass of the block= 44 N . sin α = 15 17 , sin β = 5 13 so we have cos α = 8 17 , cos β = 12 13 . The following forces are acting on the block. (1) the weight of the block 44 N directed vertically downwards, (2) The horizontal component if the tension T 1 = T 1 sin α acting towards the left side. (3) The horizontal component of the tension T 2 = T 2 sin β acting towards the right side . (4) and (5) The vertical components of the tensions T 1 cos α and T 2 cos β , acting vertically upwards, At equilibrium, Equating the horizontal forces, we have, T 1 sinα = T 2 sinβ or T 1 × 15 17 = T 2 × 5 13 or T 1 = 0 . 43 T 2 ..........(1) Equating the vertical forces, we have, T 1 cos α + T 2 cos β = 44 N or 0 . 43 × 8 17 T 2 + 12 13 T 2 = 44 Or 0 . 2 T 2 + 0 . 92 T 2 = 44 or T 2 = 39 . 2 N ......................................(2) Substituting this value in Eq.(1), we get T 1 = 0 . 43 × 39 . 2 = 16 . 9 N .....(3)

A particle has three forces acting on it, as shown in the diagram, where sin θ = 3 5 . Show that sin θ = 3 5 by resolving horizontally, and write down another equation by resolving vertically. Hence, show that G = 75 3 + 100 and find F . .

Given three forces are acting on a particle as shown in the figure Given sin θ = 3 5 therefore cos θ = 4 5 As the system is in dynamic equilibrium, the horizontal and vertical components should balance each other Resolving horizontally, we get, 240 × cos 30 = F cos θ + G cos θ 240 × 3 2 = Fcos θ + Gcos θ Or F + G cos θ = 120 3 ............................(1) Resolving the force vertically, we have F sin θ + 240 sin 30 = G sin θ Or F sin θ + 120 = G sin θ o r F - G sin θ = - 120 .......(2) Substituting the value and solving we get G = 75 3 + 100 N and F = 75 3 - 100 N

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 ° .

HARD

AS and A Level

IMPORTANT

Earn 100

A wooden block is held in position by three horizontal forces, as shown in the diagram. One acts to the left with force $56 N .$ One acts with force $F$ at an angle $θ .$ where $\sin θ = \frac{3}{5},$ , above the rightwards horizontal. One acts with force $G$ at an angle $φ,$ where $\sin φ = \frac{15}{17},$ , below the rightwards horizontal. Find $F$ and $G$ .

Important Questions on Forces in Two Dimensions

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3A>Q 11

A block with weight

44 N

is held in equilibrium by two ropes, one with tension,

T_{1},

acting at angle

\sin^{- 1} \frac{15}{17}

to the upwards vertical and the other with tension,

T_{2},

acting at angle

\sin^{- 1} \frac{5}{13}

to the upwards vertical. Find

T_{1}

and

T_{2} .

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3A>Q 12

A box with weight

400 N

is at rest on a horizontal surface. A man is pulling on a rope to try to get the box to move. The force he can exert depends on the angle at which he holds the rope, so that when the rope is at an angle

1600 \sin θ N

above the horizontal, the force he exerts is

1600 \sin θ N .

He starts by holding the rope horizontally and gradually increases the angle, thereby increasing the force. Another man tries to prevent this motion of the box, by pulling horizontally. He can exert a maximum force of

700 N .

Find the angle at which the box can no longer remain on the ground. Hence, determine whether the box lifts off the ground first or slides along the ground first.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3A>Q 13

A particle has three forces acting on it, as shown in the diagram, where $\sin θ = \frac{3}{5} .$ Show that $\sin θ = \frac{3}{5}$ by resolving horizontally, and write down another equation by resolving vertically. Hence, show that $G = 75 \sqrt{3} + 100$ and find $F .$

Question Image .

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Mechanics Course Book>Chapter 3 - Forces in Two Dimensions>EXERCISE 3A>Q 14

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 $β$ .

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EASY

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 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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EASY

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 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.

Question Image

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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Important Questions on Forces in Two Dimensions

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