Applications of forces Mied eercise 7 a Finding the components of P along each ais: ( ): P cos70 + 0sin7 ( ): P sin70 0cos7 Py tanθ P y sin70 0cos7 tanθ 0.634... cos70 + 0sin7 θ 3.6... The angle θ is 3.3 o (to 3s.f.). b Using Pythagoras theorem: P P + P P y ( cos70 0sin7 ) ( sin70 0cos7 ) P 64.9... 6.76... P has a magnitude of 6.3 N (3s.f.). a ( տ ): + + Wcosθ 40sin30 + 30sin4 ( ւ ): 30cos 4 + W sinθ 40 cos 30 Wsinθ 40cos30 30cos4 Wsinθ 40cos30 30cos4 W cosθ 40sin 30 + 30sin 4 0 3 tanθ 0.3... 0+ θ.046... The angle θ is.0 o (to 3s.f.). b Using Pythagoras theorem, W is the sum of the squares of the two components. W ( 0 3 ) ( 0 ) + + W 7... 43.34... The weight of the particle is 43.3 N (3s.f.). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
3 esolving horizontally: T cos 0 T cos0 Tcos0 T cos0 () esolving vertically: T sin 0 + T sin0 g () Tcos0 Substituting T from () into (): cos0 Tcos0 sin0 T sin 0 + g cos0 T(sin0 + cos0 tan0) g 9. T 06.6... sin0 + cos0 tan0 Subsituting this value of T into (): cos0 06.6... T 0.9... cos0 The tensions in the two parts of the rope are 06 N and 03 N (nearest whole number). 4 Let the acceleration of the system be a ms For the block, ( ր ): F ma T gsin30 a g T a+ () For the pan and masses, ( ): F ma (+ ) g T (+ ) a T 7g 7a () Since the string is inetensible, T is constant and hence () (): g a+ 7g 7a 9g a 3g a 3g Hence, the scale-pan accelerates downward, away from the pulley with magnitude a ms Applying Newton s second law to mass A, and denoting the force eerted by B on A as F AB : g F a F F F AB AB AB AB 3g g 0 g. 9.. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
4 However, Newton s third law of motion gives force eerted by B on A force eerted by A on B. Therefore the force eerted by A on B is. N. a For the kg particle ( ) T g0 T g For the kg particle: ր ( ) T+Fcosθ gsinθ 0 Fcosθ gsinθ T As T g, cosθ 4 and F 4 g 3 g F g 4 g 4 ( s.f.) sinθ 3 b ( տ ) Fsinθ gcosθ 0 Fsinθ+ gcos θ ( g 3 4 ) + ( g ) 4 9 g 4 47 ( s.f.) c F will be smaller 6 a esolving vertically: Tcos0 Tcos70 + g T(cos0 cos70) g 9. T 3.793... cos0 cos70 The tension in the string, to two significant figures, is 33 N. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 3
6 b esolving horizontally: P Tsin0 + Tsin70 P (sin 0 + sin 70) 3.793... P 4.03... The value of P is 4 N ( s.f.). 7 a ր: ( ) Tcos30 0gsin40 0 9.sin40 T cos30 363.69... The tension in the string is 364 N (3s.f.). b Even when the hill is covered in snow, there is likely to be some friction between the runners of the sled and the slope, so modelling the hill as a smooth slope is unrealistic. Let S be the reaction of the wall on the ladder at B. Let be the reaction of the ground on the ladder at A. (Both surfaces are smooth, so no friction.) ( ): F S Taking moments about A: mg a cosθ+ F a sinθ S a sinθ mg + F tanθ S tan θ (dividing by acos θ) mg + Ftanθ Ftan θ (Since F S) mg 4Ftanθ 9 9 4 F (Since tan θ ) 7. F mg F 7. mg as required. 7 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4
9 Let N be the reaction of the wall on the ladder at B. Let be the reaction of the ground on the ladder at A, Let F the friction between the ladder and the ground at A. 3 3 4 tanα sin α and cosα 4 ( ): mg+ mg 3mg Taking moments about B: mg asinα+ mg 4asinα+ F acosα asinα 3 3 mga 3 4 3 mga + + F a 6mga 3 a mga mga 3mga F a 7mga F 7mga F a 7mg The ladder and the child are in equilibrium, so F µ 7mg µ 3mg 7 µ 4 The least possible value for µ is 7 4 0 Let be the reaction of the ground on the ladder at A, Let N be the reaction of the wall on the ladder at B Let F be the friction between the wall and the ladder at B. a Since you do not know the magnitude of F, you cannot resolve vertically to find. Therefore, take moments about B (since this eliminates F): W 7a sin θ+ W a cos θ a cos θ 3 4 7W tanθ+ W (dividing through by acos θ) 7 4 4 W+ W (since tan θ ) 3 3 6W 9 W 9 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
W 0 b ( ): N 3 ( ): + F W F W W W 9 W 9 For the ladder to remain in equilibrium, F µ N W W µ 9 3 µ 3 c The ladder had negligible thickness / the ladder does not bend. a Let S be the reaction of the wall on the ladder Let be the reaction of the ground on the ladder Let F the friction between the ladder and the ground Let X be the point where the lines of action of and S intersect, as shown in the diagram. By Pythagoras s Theorem, distance from base of ladder to wall is 3 m. ( ): F S ( ): W Taking moments about X:.W 4F Suppose the ladder can rest in equilibrium in this position. Then F µ.w 0.3 W 4 3W 3W 0 30 4 which is false, therefore the assumption that F µ must be false the ladder cannot be resting in equilibrium. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
b With the brick in place, take moments about X:.W 4 F so.w 3W F 4 which is independent of M, the mass of the brick. c ( ) W +Mg ( ) F S F µ 3W +Mg) 0.3(W +Mg)3(W 0 0W W+Mg Mg W, M W 4g So the smallest value for M is W 4g Let S be the reaction of the wall on the ladder at Q Let be the reaction of the ground on the ladder at P tanα sin α and cosα 9 9 Since the ladder is in limiting equilibrium, frictional force at the wall µs 0.S. Taking moments about P: 0g cosα S 4sinα+ 0.S 4 cosα 0 4 0. g + S 9 9 9 40g.6S 39 S.4....6 ( ): F S The force F required to hold the ladder still is N ( s.f.). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
3 tanα 3 sin α 3 and cosα 4 4 u 0 ms, s 6 m, t. s, a? s ut+ at 9a 6 0+ a. 9 6 a 6 3 ( ) ւ: ( ) տ : 3gcosα F ma 6 3gsinα µ 3 3 3gsin α ( µ 3gcos α) 6 3 4 6 g µ g 3 9g µ g 0 ( ) 9 9. 0 µ 0.0697... 9. The coefficient of friction is 0.070 (3s.f.). տ : gcos30 + 0sin0 4 ( ) ր: ( ) ma F a 0 cos0 0.4 g sin 30 ( ) ( ) a 0cos0 0.4 gcos30 + 0sin0 gsin30 a 6cos0 0.4 gcos30 + 6sin0 gsin30 a 6.307... The acceleration of the particle is 6.3 ms (3s.f.). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
Since m > µ m, when system is released from rest then B moves downwards and A moves towards the pulley P For particle A: : µ mg ( ) ( ): F ma T µ ma T µ mg ma For particle B: ( ): T ma+ µ mg () F ma mg T ma T mg ma () Since string is inetensible, the values of T are identical, so () (): ma+ µ mg mg ma ma+ ma mg µ mg ( µ ) g m m a m + m as required. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9
6 For particle with mass m: ր ( ) F ma m T mg sin30 T + gsin30 m () For particle with mass m : ր ( ) F ma m mg sin4 T gsin4 m T () Since string is inetensible, T is constant throughout and hence () (): + gsin30 m gsin4 m Challenge g g + m m m g as required m + g Since the rod is uniform, the weight acts from the midpoint of AB. a Take moments about A: ( ) ( ) 0g. + g T 3sin60 g T 3sin 60 9. T 94.300... 3 3 The tension in the string is 94.3 N (3s.f.). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 0
Challenge b Let the horizontal and vertical components of the reaction at the hinge be and y respectively. esolving horizontally: Tcos60 g cos60 3sin 60 g g 3tan60 3 3 esolving vertically upwards: Tsin60 0g g y y y g sin60 g 3sin 60 0g g 3 3 The reaction at the hinge is given by: + y g 0g + 3 3 3 6 400 + g 7 9 9. 0.70... 7 y tanθ 0g 3 3 4 3 tanθ 3 g θ 4... The reaction at the hinge is 0.6 N acting at 4. o below the horizontal (both values to 3 s.f.). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.