A-Level Topic: Resultant Forces 7 Starter Activity Chapter ... · Topic: Motion in 2 Directions . Chapter Reference: Mechanics 1, Chapter 10 . 7 . minutes . Solutions 1a. Resolving

1. Two perpendicular forces of magnitudes F N and 8N act at a point O (see diagram). Their resultant has magnitude 17 N. a. Calculate F and find the angle which the resultant makes with the 8 N force. (4) A third force of magnitude E N, acting in the same plane as the two original forces, is now applied at the point O. The three forces of magnitudes E N, F N and 8 N are in equilibrium. b. State the value of E and the angle between the directions of the E N and 8 N forces. (2) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Resultant Forces Chapter Reference: Mechanics 1, Chapter 10

7 minutes

Solutions

1a. F2 = 172 – 82 M1 F = 15 M1 cos x = 8

17 M1

x = 61.9o M1 1b.

E = 17 M1 Angle = 118.1o or 241.9o M1

1. A particle P of mass 2 kg is attached to one end of a light string, the other end of which is attached to a fixed point O. The particle is held in equilibrium, with OP at 30º to the downward vertical, by a force of magnitude F newtons. The force acts in the same vertical plane as the string and acts at an angle of 30º to the horizontal, as shown in the figure below. Find, the value of F and the tension in the string. (7) __________________________________________________________________________________________

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Topic: Motion in 2 Directions Chapter Reference: Mechanics 1, Chapter 10

8 minutes

Solutions 1. Resolving Vertically: M1 T cos 30 + F cos 60 = 2g M1 Resolving Horizontally: M1 T cos 60 – F cos 30 = 0 M1 Solving simultaneously/elimination M1 F = g = 9.8N M1 T = √3𝑔𝑔 = 17 N M1

1.A particle of mass m kg is attached at C to two light inextensible strings AC and BC. The other ends of the strings are attached to fixed points A and B on a horizontal ceiling. The particle hangs in equilibrium with AC and BC inclined to the horizontal at 30° and 60° respectively, as shown in the figure. Given that the tension in AC is 20 N, find, a. The tension in BC (4) b. The value of m (4) __________________________________________________________________________________________

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8 minutes

Solutions

1a. Resolving horizontally M1 20 cos 30 = M1 20 cos 30 = T cos 60 M1 T = 20√3 N M1

1b.

Resolving vertically, M1 mg = M1 mg = 20 sin 30 + T sin 60 M1 m = 40

𝑔𝑔= 4.08 𝑘𝑘𝑘𝑘 M1

1. Two forces P and Q act on a particle O. The angle between the lines of action of P and Q is 120o as shown in the figure below. The force P has magnitude 20 N and the force Q has magnitude X newtons. The results of P and Q is the for R. Given that the magnitude of R is 3X newtons, find, giving your answers to 3 significant figures. a. The value of X (4) b. The magnitude of (P – Q) (3) __________________________________________________________________________________________

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Topic: Resultant Forces (D) Chapter Reference: Mechanics 1, Chapter 10

8 minutes

Solutions

1a. Use of the cosine rule: (3X)2 = 202 + X2 – 2.20Xcos 60 M1

8X2 + 20X – 400 = 0 M1 X = −5± √25+800

4 M1

X = 5.93 M1 1b.

Use of the cosine rule M1 |𝑃𝑃 − 𝑄𝑄|2 = 202 + X2 – 2X x 20 x cos 120 M1 |𝑃𝑃 − 𝑄𝑄| = 23.5 N M1

1. A particle of weight W newtons is attached at C to two light inextensible strings AC and BC. The other ends of the strings are attached to fixed points A and B on a horizontal ceiling. The particle hangs in equilibrium with AC and BC inclined to the horizontal at 30° and 50° respectively, as shown in the figure. Given that the tension in BC is 6N, find, a. The tension in AC (3) b. The value of W (3) __________________________________________________________________________________________

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7 minutes

Solutions

1a. Resolving horizontally M1 T cos 30 = 6 cos 50 M1 T = 4.45 N (No units, no mark) M1

1b.

Resolving vertically M1 W = 6 cos 40 + T cos 60 M1 W = 6.82 N (No units, no mark) M1

1. Two forces F1 and F2 act on a particle P. The force F1 is given by F1 = (-i + 2j)N and F2 acts in the direction of the vector (i + j). Given that the resultant of F1 and F2 actions in the direction of the vector (i + 3j). Find F2. (6) 2. Three forces (15i + j)N, (5qi – pj)N and (-3pi – qj)N, where p and q are constants, act on a particle. Given that the particle is in equilibrium, find the value of p and the value of q. (5) __________________________________________________________________________________________

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Topic: Forces and Vectors Chapter Reference: Mechanics 1, Chapter 10

8 minutes

Solutions

1. F1 + F2 = F3 M1 (-1 + k)i + (2 + k)j = ai + 3aj M1 Equating i and j values -1 + k = a 2 + k = 3a

M1

Solving simultaneously: -3 + 3k = 2 + k 2k = 5

M1

k = 2.5 M1 F2 = 2.5i + 2.5j M1

2.

15 + 5q – 3p = 0 M1 Equating vertical forces, 1 – p – q = 0 p = 1 – q

M1

Solving by substitution, 15 + 5q – 3(1 – q) = 0 M1

12 + 8q = 0 q = -3

2 M1

p = 52 M1

1. A vertical rope AB has its end B attached to the top of the scale plan. The scale pan has mass 0.5kg and carries a brick of mass 1.5kg, as shown in the figure. The scale pan is raised vertically upwards with constant acceleration 0.5 ms-2 using the rope AB. The rope is modelled as a light inextensible string. a. Find the tension in the rope AB. (4) b. Find the magnitude of the force exerted on the scale pan by the brick (4) __________________________________________________________________________________________

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Topic: Forces and Acceleration (scale-pan) Chapter Reference: Mechanics 1, Chapter 10

6 minutes

Solutions

1a. Using F = ma in the vertical direction M1 T – W = (0.5 + 1.5) x 0.5 T – 2g = 2 x 0.5 M1

T = 2g + 1 M1 T = 20.6 N M1

1b.

Using F = ma in the vertical direction M1 R – W = ma R = ma + W M1

R = 1.5 x 0.5 + 1.5 x 9.8 M1 R = 15.45 N M1

1. A particle P of mass 0.5 kg is moving under the action of a single force (3i – 2j)N. Show that the magnitude of the acceleration of P is 2√13 ms-2. (5) 2. Two cars P and Q are moving on a straight horizontal roads with constant velocities. The velocity of P is (15i + 20j) ms-1 and the velocity of Q is (20i – 5j)ms-1. Find the direction of motion of Q, giving your answer as a bearing to the nearest degree. (3) __________________________________________________________________________________________

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Topic: Forces, Acceleration and Vectors Chapter Reference: Mechanics 1, Chapter 10

5 minutes

Solutions

1. Use of F = ma M1 (3i – 2j) = 0.5 x a M1 a = 6i - 4j M1 a = √62 + 42 M1 a = 2√13 ms-2 M1

2.

Bearing of Q = 90 + x M1 Q = 90 + tan-1( 5

20) M1

Q = 90 + 14.0 Q = 104o M1

1. Two particles A and B have masses 2m and 3m respectively. The particles are attached to the ends of a light inextensible string. Particle A is held at rest on a smooth horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table. Particle B hangs at rest vertically below the pulley with the string taut, as shown in the figure. Particle A is released from rest. Assuming that A has not reached the pulley, find, a. the acceleration of B (6) b. the tension in the string (1) c. the magnitude and direction of the force exerted on the pulley by the string (2) __________________________________________________________________________________________

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Topic: Forces and Acceleration (Connected Particles) Chapter Reference: Mechanics 1, Chapter 10

8 minutes

Solutions

1a. Using F = ma on A M1 T = 2ma M1 Using F = ma on B M1 3mg – T = 3ma M1 Solving simultaneously, 3mg = 5ma M1

a = 3𝑔𝑔5

(or 5.88) ms-2 M1 1b.

Tension = 6𝑚𝑚𝑔𝑔5

(11.76) N M1 1c.

F = √𝑇𝑇2 + 𝑇𝑇2 = 6𝑚𝑚𝑔𝑔

5 √2 N M1 tan 𝜃𝜃 = 1 𝜃𝜃 = 45o M1

1. Two particles A and B have masses 2m and 3m respectively. The particles are connected by a light inextensible string which passes over a smooth light fixed pulley. The system is held at rest with the string taut. The hanging parts of the string are vertical and A and B are above a horizontal plane, as shown in Figure 2. The system is released from rest. Show that the tension in the string immediately after the particles are released is 12

5𝑚𝑚𝑚𝑚. (8)

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Topic: Forces and Acceleration (Connected Particles) Chapter Reference: Mechanics 1, Chapter 10

6 minutes

Solutions

1. F = ma for particle B M1 3mg - T = 3ma M1 F = ma for particle A M1 T – 2mg = 2ma M1 Solving simultaneously, 2mg – T + T – 2mg = 3ma + 2ma M1

mg = 5ma a = 𝑔𝑔

5 M1

T – 2mg = 2m(𝑔𝑔5

) T = 2m(𝑔𝑔

5+ 𝑚𝑚)

M1

T = 125𝑚𝑚𝑚𝑚 M1

1. A vertical light rod PQ has a particle of mass 0.5kg attached to it at P and a particle of mass 0.75kg attached to it at Q, to form a system, as shown in the figure below. The system is accelerated vertically upwards by a vertical force of magnitude 15N applied to the particle at Q. Find the thrust in the rod. (8) __________________________________________________________________________________________

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Topic: Forces and Acceleration (Connected Particles with Thrust) Chapter Reference: Mechanics 1, Chapter 10

8 minutes

Solutions

1. Using F = ma at P M1 T – 0.5g = 0.5a M1 Using F = ma at Q M1 15 – T – 0.75g = 0.75a M1 Solving simultaneously, 15 – 1.25g = 1.25a

M1

a = 2.2 ms-2 M1 Hence, T = 0.5 x 2.2 + 0.5g M1 T = 6N M1

1. A woman travels in a lift. The mass of the woman is 50 kg and the mass of the lift is 950 kg. The lift is being raised vertically by a vertical cable which is attached to the top of the lift. The lift is moving upwards and has constant deceleration of 2 ms-2. By modelling the cable as being light and inextensible find. a. The tension in the cable (3) b. The magnitude of the force exerted on the woman by the floor of the lift (3) __________________________________________________________________________________________

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Topic: Force and Acceleration (Lifts) Chapter Reference: Mechanics 1, Chapter 10

6 minutes

Solutions

1a.

F = ma M1 T – 1000g = 1000 x -2 M1 T = 7810 N M1

1b.

Resolving vertically, R – 50g M1

Using F = ma R – 50g = 50 x – 2 M1

R = 390 N M1

Mech 1 C10 Resultant Forces 1Mech 1 C10 Resultant Forces 2Mech 1 C10 Resultant Forces 3Mech 1 C10 Resultant Forces (Difficult)Mech 1 C10 Motion in 2 DirectionsMech 1 C10 Forces and VectorsMech 1 C10 Forces and Acceleration (Scale Pan)Mech 1 C10 Forces, Acceleration and VectorsMech 1 C10 Connected ParticlesMech 1 C10 Connected Particles (Pulleys)Mech 1 C10 Connected Particles with ThrustMech 1 C10 Forces and Acceleration (Lifts)

Documents

A-Level Topic: Resultant Forces 7 Starter Activity Chapter ... · Topic: Motion in 2 Directions . Chapter Reference: Mechanics 1, Chapter 10 . 7 . minutes . Solutions 1a. Resolving