Applications of forces 7D Suppose that the rod has length a. Taking moments about A: at 80 acos0 T 80 T 0. 6 N R( ), F T sin0 0 7. N R, T cos0 + R 80 R 80 0 50N In order for the rod to remain in equilibrium, we must have F R: 0 50 0 50 5 minimum 0.5 ( s.f.) Let A be the end of the ladder on the ground. Let F be the frictional force at A. a Taking moments about A: 0g.5cos 65 S 5sin 65 5g cos 65 S 5sin 65 5g tan65.8n b R( ), F S.8N R, R 0g 98N c To ensure ladder remains in equilibrium, we must have F R.8 98 0. ( s.f.) d The weight is shown as acting through the midpoint of the ladder because of the assumption that the ladder is uniform. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
Let the ladder have length a, and be inclined at θ to the horizontal. a R, R 0g Taking moments about A: 0g a+ F asinθ R a 0g + Fsinθ 60gcos θ (using R 0 g) F sinθ 0g 0g F The ladder is on the point of slipping, so F R 0g 0 g 8 9 θ.6 b R, R 0g R, N F 0 N F Taking moments about B: 0g a N asinθ 0g a F asinθ 0g F sinθ 0g F The ladder is on the point of slipping, so F R 0g 0 g 9 θ.0 N is the normal reaction at A, R is the normal reaction at B, F is the frictional force at B. c The assumption that the wall is smooth means there is no friction between the ladder and the wall. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
a Suppose that the boy reaches the point B, a distance x from A, whilst the end of the ladder is still in contact with the ground. R( ), F N R, R 50g Taking moments about A: 0g + 0g x N 8sinθ 0gx 8N 0gx N 8 0gx N (since ) 6 0gx F (since F N) 6 0gx R (in limiting equilibrium) 6 0gx 0. 50g 6 0 80+ 0x x 5 m b i The ladder may not be uniform. ii There would be friction between the ladder and the wall. 5 Let: S be the normal reaction of the rail on the pole at C, R be the normal reaction of the ground on the pole at A, F be the friction between the pole and the ground at A. θ be the angle between the pole and the ground. From the diagram, sinθ.5 and hence 9 5 a Taking moments about A:.5S 5 5 8 5 S N 9 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
5 b R( ) F S sinθ 8 5 9 R( ) 6 5 7 R+ S 8 5 5 R 9 0 7 68 7 Pole is in limiting equilibrium, so F R 6 5 68 7 7 6 5 68 5 7 0.56 ( s.f.) 5 c The assumption that the rail is smooth means there is no friction between the rail and the pole. 6 Suppose that the ladder has length a and weight. Let: S be the normal reaction of the wall on the ladder, R be the normal reaction of the floor on the ladder, F be the friction between the floor and the ladder. X be the point where the lines of action of and S meet. Taking moments about X: asinθ F R a F sinθ R () The ladder is in limiting equilibrium, so F R Substituting F R in (): Rsinθ R sinθ sinθ Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
7 Let: N be the normal reaction of the drum on the ladder at P, R be the normal reaction of the ground on the ladder at A, F be the friction between the ground and the ladder at A. Taking moments about A: 0g.5cos 5 5N R( ) N cos 5 + R 0g R( ) F N sin5 0g.5cos 5 N 5 g cos 5 R 0g g cos 5 cos 5 0.9...N g cos 5 sin 5 6.6...N F R to maintain equilibrium: g cos5 sin5 (0g g cos 5 ) cos5 sin5 0 cos 5 0.60 ( s.f.) Least possible is 0.60 ( s.f.) 8 Let: R be the reaction of the ground on the ladder F be the friction between the ground and the ladder S be the reaction of the wall on the ladder G be the friction between the wall and the ladder. X be the point where the lines of action R and S meet. Suppose that the ladder has length a and weight. As the ladder rests in limiting equilibrium,f R and G S. Taking moments about X: a F asinθ + G a R( ), F + G () F S R, R+ G Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5
8 Substituting for and F in equation (): R+ G R + G R G R R R R Hence (Since G S F R ) 9 Let: A and B be the ends of the ladder. P be the normal reaction of the wall on the ladder at B, R the normal reaction of the ground on the ladder at A F be the friction at between the ladder and the ground at A Let the length of the ladder be a. a Taking moments about A: a cos 60 P a cos 0 acos60 P a cos 0 P P () b R( ), R () R( ), F P () Now F R since the ladder is in equilibrium (if not, ladder would slide) Hence, PR (by ) R 6 (by ) (by ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
9 c Let: R be the normal reaction of the ground on the ladder at A P be the normal reaction of the wall on the ladder at B, l be the length of the ladder Since the ladder is in limiting equilibrium, F ' R' R( ), R + w R( ), R P Taking moments about B: l cos60 + F lsin60 R l cos60 R + R + w + (since ' and ) + w 5 R + w 5 6 + ( + w ) ( + w ) 5 5 + 6 + 6w 0 + 0w w w 0 Let: T be the normal force of the peg on the rod at P, G be the frictional force at P, S be the normal force of the peg on the rod at Q, F be the frictional force at Q. a Taking moments about P: S 0 0 5 cos 0 S 0 5 0 5 S N Taking moments about Q: T 0 0 5 cos 0 T T 0 5 0 5 N Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
0 b Rց G+ F 0cos60 0 () Since the rod is about to slip, friction is limiting and hence G T, F S. From part a, 0 G+ F T+ S 0 () ( ) a Let: S be the normal reaction of the wall on the ladder at Y, R be the normal reaction of the ground on the ladder at X F be the friction at between the ladder and the ground at X so sin θ and Ladder is in equilibrium. Taking moments about X: l + 9l Slsinθ 9 S + S + 9 9 S 9 S b R( ): R + 9 0 For the ladder to be in limiting equilibrium, F R F 0 5 F R( ): If P + F > S, ladder will slide towards and up the wall If P < S F, ladder will slide away from and down the wall Therefore S F P S + F Substituting values for S & F from part a and above: 9 9 P + 9 9 P + Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8
c Modelling the ladder as uniform allows us to assume the weight acts through the midpoint. d i The reaction of the wall on the ladder will decrease. To understand why, consider how we took moments about X in part a l + 9l Slsinθ The first term in this equation is the turning moment of the weight of the ladder, which acts at a l l distance from X. If the centre of mass of the ladder is more towards X, say where a>, a then this first term would decrease and hence S would also decrease. ii Ladder remains in equilibrium when S F P S + F If S were to decrease, then this range of values for P would also decrease. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9