Applications of forces 7C

Transcription

1 Applications of forces 7C 1 Let the normal reaction be R N, the friction force be F N and the coefficient of friction be µ. Resolve horizontally to find F, vertically to find R and use F µr to find µ: 8cos2 F F 8cos2 R( ) R+8sin2 2g R2g 8sin2 As the book is on the point of slipping the friction is limiting: F µr µ F R 8cos 2 2g 8sin (3 s.f.) 2 Let the normal reaction be R N, the friction force be F N and the coefficient of friction be µ. : 6cos3 F F 6cos (3 s.f.) R( ): R 6sin3 4g R 6sin3+ 4 g As the block is on the point of slipping F µ R F µ R.123 (3 s.f.) Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 3 Let the normal reaction force be R and the friction force be F. a Resolve horizontally to find the magnitude of the friction force necessary to maintain equilibrium: 3cos 6 F F 3cos6 F 1.5N b Resolve vertically to calculate R and henceµ R: ( ) R R+ 3sin6 1 R 1 3sin (3 s.f.) µr (3 s.f.) Since F 1.5N< 2.2N µ R, the friction required to maintain equilibrium is not limiting friction. 4 a Let the normal reaction be R N and the friction force required to maintain equilibrium be F N. Let the mass of the books be m kg. 147 F F 147N R( ) R 1g mg R1g+mg As the equilibrium is limiting, F µr 147.3(1 g+ mg) 147 3g+.3 mg 147 3g m.3g 4kg b The assumption is that the crate and books may be modelled as a particle. Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 2

3 5 a Let R be the normal reaction and F be the force of friction when P acts downwards. Pcos45 F FPcos45 R( ) R Psin45 2g RPsin45+2g Resolve horizontally and vertically to find F and R, then use the condition for limiting friction. As the equilibrium is limiting, F µr Pcos45.3( Psin45+ 2 g) P(cos45.3sin45 ).6g.6g P cos45.3sin45 6g N (3 s.f.) b Let R ' be the normal reaction and F ' be the force of friction when P acts upwards. Pcos45 F' ( ) R F' Pcos 45 R' + Psin45 2g R' 2g Psin45 As the equilibrium is limiting, F µr Pcos45.3(2g Psin45 ) P(cos45+.3sin45 ).6g 6g 2 P N (3 s.f.) Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 6 Let R be the normal reaction and F be the force of friction required to maintain equilibrium. Since the particle is on the point of slipping up the plane, the force of friction acts down the slope. 3 F.3gsin3 R( տ ) R.3gcos3 F 3.3gsin3 1.53N R.3gcos N As the particle is on the point of slipping, F µ R 1.53 µ µ (3 s.f.) (accept.6) 7 Let R be the normal reaction and F be the force of friction required to maintain equilibrium. Since the particle is on the point of slipping up the plane, the force of friction acts down the slope. a R( տ ) R 1.5gcos25 R 1.5gcos25 b R 1.5gcos25 X F 1.5gsin N (3 s.f.) X F+ 1.5gsin25 (1) The particle is in limiting equilibrium, so F µ R F (using R 13.3 N from a) Sub F 3.33 N into (1): X gsin N (3 s.f.) Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 8 Let the normal reaction be R and the friction force be F acting down the plane. a R( տ ) R 2sin3 1.5gcos3 R 2sin3+ 1.5gcos (3 s.f.) The normal reaction has magnitude 22.7 N or 23 N (2 s.f.). b 2cos3 F 1.5gsin3 F 2cos3 1.5gsin (3 s.f.) The friction force has magnitude 9.97 N and acts down the plane. c Minimum possible value of µ occurs when frictional force required to maintain equilibrium is µ R: F µ R µ (using F from band Rfrom a) µ If you are told the particle is in equilibrium, but not told which way the particle is about to slip, then draw a diagram showing all the forces acting on the particle, with friction acting down the plane. Resolve forces parallel to the plane. If F > then you have chosen the correct direction. If F < then you know friction acts up the plane. The coefficient of friction must be at least.439 (3s.f.) to prevent the block sliding. Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 9 Let the normal reaction be R and the friction force be F acting down the plane. a R( տ) R Xsin4 3gcos4 R Xsin4+ 3gcos4 (1) Xcos4 F 3gsin4 F Xcos4 3gsin4 (2) As the friction is limiting, F µ R Using F from (2) and R from (1) gives: Xcos4 3gsin4.3( Xsin4+ 3gcos4 ) Xcos4.3Xsin4.9gcos4+ 3gsin4 X(cos4.3sin4 ).9gcos4+ 3gsin4.9gcos4+ 3gsin4 X cos4.3sin X 44.8 N (3 s.f.) b Substituting X 44.8 Ninto equation (1) gives R 44.8 sin4+ 3gcos N (3 s.f.) Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 1 Let the normal reaction be R and the friction force be F acting up the plane. The friction acts up the plane, as the sledge is on the point of slipping down the plane. T+ F 22gsin35 (1) R( տ) R 22gcos35 R 22gcos35 R N As the friction is limiting, F µ R F N (3 s.f.) Substituting T 22.1 N into equation (1) gives: T 22g sin N (3 s.f.) 11 R( տ ) R.5gcos4 + Tsin2 R.5g cos 4 T sin 2 T occurs when the particle is on the point of moving up the plane. At this point, limiting friction F µ R acts down the plane: T cos2.5gsin4 F 1 Tcos2.5gsin4 (.5gcos4 Tsin2 ) 5 T cos2 +.2T sin2.5gsin4 +.1gcos4 T.5gsin4 +.1gcos4 cos2 +.2sin Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 11 T MIN occurs when the particle is on the point of moving down the plane. At this point, limiting friction F µ R acts up the plane: T cos2.5gsin4 F MIN 1 TMINcos2.5gsin4 + (.5gcos4 Tsin2 ) 5 T cos2.2t sin2.5gsin4.1gcos4 MIN MIN T T lies between 2.75 N and 3.87 N (both values to 3s.f.)..5gsin4.1gcos4 cos2.2sin R( տ ) R+ 1sin2 gcos4 1cos2 F gsin4 R gcos4 1sin F 1cos2 gsin As the friction is limiting, F µ R F µ R (3 s.f.) Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 13 tan θ 3 so sin θ 3 and cosθ Before P is applied, the particle will be on the point of moving down the slope and the limiting frictional force therefore acts up the slope: R( տ ) R 2gcosθ 2gsinθ F Max 2gsinθ µ R 2gsinθ µ 2gcosθ µ tanθ µ 3 4 After applying the max value of P that will allow the particle to remain in equilibrium, particle is on the point of moving up the slope. Therefore, the frictional force acts down the slope. R( տ ) R 2gcosθ + Psinθ Pcosθ F + 2gsinθ Max Pcosθ µ R + 2gsinθ Pcos θ µ (2gcosθ+ Psin θ) + 2gsinθ ( ) P(cosθ µ sin θ ) 2g µ cosθ + sinθ P P So max. P is 67.2 N P 67.2 N Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 9