7.2 Wedges Example 1, page 1 of 4
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1 7.2 Wedges
2 7.2 Wedges Example 1, page 1 of 4 1. If the coefficient of static friction equals 0.3 for all surfaces of contact, determine the smallest value of force P necessary to raise the block. Neglect the weight of the wedge B. 300 kg 10 B P
3 7.2 Wedges Example 1, page 2 of Impending motion of left surface of block relative to wall Free-body diagram of block y 6 + Equations of equilibrium for block Fx = 0: N f B cos 10 N B sin = 0 (1) F y = 0: f f B sin 10 N B cos kn = 0 (2) Impending motion so, f = f -max N = 0.3N (3) f B = f B-max N B = 0.3N B (4) N f (300 kg)(9.81 m/s 2 ) = kn f B x 4 Impending motion of lower surface of block relative to block B (This is the motion that an observer sitting on block B would see as he observes block move past.) 3 The friction force from the wall opposes the relative motion of block B. 10 N B 5 The friction force from block B opposes the relative motion of block.
4 7.2 Wedges Example 1, page 3 of 4 7 Geometry 9 Free-body diagram of wedge B Impending motion of top surface of block B relative to block. 10 B N B = N f B = N P 10 8 = = 10 Using = 10 in Eqs. 1-4 and solving simultaneously gives f = N N = N 12 Impending motion of lower surface of block B relative to floor N B f B The friction force f B from block opposes the relative motion of block B. The friction force f B from the floor opposes the relative motion of block B. f B = N N B = N 14 Equations of equilibrium for block B + Fx = 0: f B + (1.115 N) cos 10 + (3.716 N) sin 10 P = 0 (5) F y = 0: N B + (1.115 N) sin 10 (3.716 N) cos 10 = 0 (6)
5 7.2 Wedges Example 1, page 4 of 4 15 Slip impends between block B and the floor, so f B = f B-max N B = 0.3N B (7) Solving Eqs. 5, 6, and 7 simultaneously gives f B = 1.04 N N B = 3.47 N P = 2.78 N ns.
6 7.2 Wedges Example 2, page 1 of 3 2. Wedges and B are to be glued together. Determine the minimum coefficient of static friction required, if clamp CDE is to be able to hold the wedges in the position shown, while the glue dries. C D B E
7 7.2 Wedges Example 2, page 2 of 3 3 The friction force from the upper jaw of the clamp opposes the relative motion of block. 1 f Free-body diagram of wedges N 10 2 Impending motion of upper surface of wedge relative to the upper jaw of the clamp (The wedge is just about to slip out from the jaws of the clamp.) 5 The friction force from the lower jaw of the clamp opposes the relative motion of block B. f N 10 B 4 Impending motion of lower surface of wedge B relative to the lower jaw of the clamp Because of symmetry, the normal force N and friction force f acting on the bottom of the wedge are given the same variable names as those on the top. Equations of equilibrium Fx = 0: 2N sin 2f cos 10 = 0 (1) F y = 0: N cos N cos f sin 10 + f sin 10 = 0 (2)
8 7.2 Wedges Example 2, page 3 of Since we have already used symmetry in labeling the forces on the free-body diagram, the F y equilibrium equation degenerates to 0 = 0 and gives us no new information. Geometry 11 Eqs. 1 and 3 constitute two equations in three unknowns (f, N, and ), so it appears that we can't solve the problem. But we have not been asked to find f and N, only. Thus solve Eq. 1 with = 10 for the ratio f/n: f N = tan 10 (4) = = lso solve Eq. 3 for f/n: f N = (5) 10 Slip impends between both jaws of the clamp and the wedge, and thus f = f max N (3) Eqs. 4 and 5 imply that = tan 10 = ns. This result is independent of the clamp force N. That is, no matter how strong the clamp spring is, if the coefficient of friction is less than 0.176, then the clamp will slip.
9 + 7.2 Wedges Example 3, page 1 of 3 3. Determine the smallest values of forces P 1 and P 2 required to raise block while preventing from moving horizontally. The coefficient of static friction for all surfaces of contact is 0.3, and the weight of wedges B and C is negligible compared to the weight of block. 1 Free-body diagram of block 2 kip 2 kip f B N B 7 P 2 B P 1 C 2 Equations of equilibrium for block Fx = 0: f B = 0 (1) Therefore, f B = 0 F y = 0: N B 2 kip = 0 (2) Solving gives N B = 2 kip 3 The friction force f B has to be zero, since we know that block is not to move horizontally and no other horizontal force acts. In fact, we could have just shown f B = 0 on the free body initially.
10 7.2 Wedges Example 3, page 2 of 3 4 Free-body diagram of block B y 9 Geometry N B = 2 kip B P 2 x 90 7 = 83 f BC 6 N BC The friction force from block C opposes the relative motion of block B. 5 Impending motion of lower surface of block B relative to block C. = = 7 7 Equations of equilibrium for block B 10 Solving Eqs. 3, 4, and 5 with = 7 gives + Fx = 0: f BC cos 7 + N BC sin P 2 = 0 (3) N BC = kip F y = 0: f BC sin 7 + N BC cos 2 kip = 0 (4) f BC = kip = 628 lb 8 Slip impends between blocks B and C, so P 2 = kip = 878 lb ns. f BC = f BC-max N BC = 0.3N BC (5)
11 7.2 Wedges Example 3, page 3 of 3 11 Free-body diagram of block C = 7 y N BC = kip 13 The friction force f BC from block B opposes the relative motion of block C. P 1 C 7 f BC 12 Impending motion of upper surface of C relative to B x 15 The friction force f C from the floor opposes the relative motion of block B. f C N C 14 Impending motion of lower surface of C relative to the floor 16 Equilibrium equations for block C + F x = 0: f BC cos 7 (2.092 kip) sin 7 + P 1 f C = 0 (6) F y = 0: f BC sin 7 (2.092 kip) cos 7 + N C = 0 (7) Slip impends between block C and the floor, so f C = f C-max N C = 0.3N C (8) 17 Solving Eqs. 6, 7, and 8 simultaneously gives f C = 0.6 kip = 600 lb N C = 2 kip P 1 = kip ns.
12 7.2 Wedges Example 4, page 1 of 4 4. If the coefficient of static friction for all surfaces of contact is 0.25, determine the smallest value of the forces P that will move wedge B upward. B 200 kg P 20 kg kg C P
13 7.2 Wedges Example 4, page 2 of 4 1 Free-body diagram of block Weight = (20 kg)(9.81 m/s 2 ) = N 4 Impending motion of right-side of block relative to block B (n observer on block B would see block move down.) P 75 f B N B 5 The friction force from block B opposes the relative motion of block. 2 Impending motion of bottom of block relative to ground f 3 The friction force from the floor opposes the motion of block. N 6 Equations of equilibrium + Fx = 0: P f f B cos 75 N B cos = 0 (1) 7 Geometry F y = 0: N N + f B sin 75 N B sin = 0 (2) Slip impends so, f = f -max N = 0.25N (3) = = 15 f B = f B-max N B = 0.25N B (4)
14 7.2 Wedges Example 4, page 3 of 4 8 Free-body diagram of block B Impending motion of block B relative to block N B f B = B Weight = (200 kg)(9.81 m/s 2 ) = 1962 N 9 = 15 Impending motion of block B relative to block C The friction forces from blocks and C oppose impending upward relative motion of block B. f BC = f B by symmetry (You can show this by summing moments about the point where the lines of action of N B, N BC, and the weight intersect) N BC = N B by symmetry (You can show this by summing moments about the point where the lines of action of f B, f BC, and the weight intersect.) 10 + Equations of equilibrium Fx = 0: N B cos 15 N B cos 15 + f B cos 75 f B cos 75 = 0 (5) (Note that this equation reduces to 0 = 0. This happens because we have assumed symmetry to conclude that f BC = f B and N BC = N B.) F y = 0: N B sin 15 + N B sin 15 f B sin 75 f B sin N = 0 (6)
15 7.2 Wedges Example 4, page 4 of 4 11 Solving Eqs. 1 4 and 6 simultaneously, with = 75, gives f = 294 N = kn N = N = 1.18 kn f B = N = kn N B = N = 56.6 kn P = N = 58.6 kn ns.
16 7.2 Wedges Example 5, page 1 of 4 5. The cylinder D, which is connected by a pin at to the triangular plate C, is being raised by the wedge B. Neglecting the weight of the wedge and the plate, determine the minimum force P necessary to raise the cylinder if the coefficient of static friction is 0.3 for the surfaces of contact of the wedge. 400 lb D C r = 10 in 10 B P
17 7.2 Wedges Example 5, page 2 of 4 1 Free-body diagram of triangular plate y 2 Equilibrium equation for triangular plate F y = 0: y = 0 (1) y So, no vertical force is transmitted by pin. Resultant of roller forces x x C
18 + 7.2 Wedges Example 5, page 3 of 4 3 Free-body diagram of cylinder D 4 Equilibrium equations for cylinder D. r = 10 in. x 400 lb D The equation for the sum of forces in the x-direction has not been included because including it would introduce an additional unknown, x, which we have not been asked to determine. F y = 0: f BD sin + N BD cos 400 lb = 0 (2) M = 0: f BD (10 in.) = 0 (3) y = 0 Therefore f BD = 0, that is, no friction force acts on the cylinder. 10 f BD N BD 5 Geometry = = 10 6 Solving Eq. 2 with = 10 gives N BD = lb
19 7.2 Wedges Example 5, page 4 of 4 7 Free-body diagram of wedge B 9 Equilibrium equations for wedge B N BD = lb + Fx = 0: f B + (406.2 lb) sin 10 P = 0 (4) = 10 B P F y = 0: N B (406.2 lb) cos 10 = 0 (5) Slip impends, so 10 f B = f B-max N B = 0.3N B (6) 8 Impending motion of wedge B relative to the floor. N B f B Solving Eqs. 4, 5, and 6 simultaneously gives f B = 120 lb N B = 400 lb P = lb ns.
20 7.2 Wedges Example 6, page 1 of 3 6. To split the log shown, a 120-lb force is applied to the top of the wedge, which causes the wedge to be about to slip farther into the log. Determine the friction and normal forces acting on the sides of the wedge, if the coefficient of static friction is 0.6. lso determine if the wedge will pop out of the log if the force is removed. Neglect the weight of the wedge. 120 lb Wedge angle = 8
21 7.2 Wedges Example 6, page 2 of 3 1 Free-body diagram of wedge 120 lb 5 Equilibrium equation for the wedge F y = 0: 2f cos 4 2N sin 120 lb = 0 (1) 2 Impending motion of left side of wedge 6 Geometry N The friction force from the wood opposes the motion of the wedge. 4 f f N Because of symmetry, the same variables, N and f, are used on the right side of the wedge = = = 4 Slip impends, so f = f max N = 0.6N (2) Solving Eqs. 1 and 2 simultaneously gives f = 53.9 lb N = 89.8 lb ns. ns.
22 7.2 Wedges Example 6, page 3 of 3 8 Second part of the problem (Determine if the wedge will pop out, if no force acts down on the top). To determine if the wedge will pop out, let's first determine what force would be needed to pull the wedge out. 9 Free-body diagram of wedge based on assumption that a force P is applied to pull the wedge out. P 12 Equilibrium equation for the wedge 10 Impending motion of wedge (The wedge is just about to pop out.) f f F y = 0: 2f ' cos 4 + 2N' sin 4 + P = 0 (3) Slip impends, so f ' = f 'max N' = 0.6N' (4) N N Substituting the expression for f ' of Eq. 4 into Eq. 3 and solving for P gives 11 The friction force from the wood opposes the motion of the wedge (The force tries to keep the wedge in the stump.) P = 2(0.6 cos 4 sin 4 )N' = N' (5) Since the normal force N' always points towards the wedge, it is always positive. Eq. 5 thus implies that the wedge will pop out only if an upward force greater than or equal to times the normal force is applied. For any smaller value of P, the wedge will remain in place. Thus in particular for the special case of P = 0 (no vertical force applied), the wedge will remain in place.
23 7.2 Wedges Example 7, page 1 of 6 7. The end of the beam needs to be raised slightly to make it level. If the coefficient of static friction of the contact surfaces of the wedge is 0.3, determine the smallest value of the horizontal force P that will raise end. The weight and size of the wedge are negligible. lso, if the force P is removed, determine if the wedge will remain in place, that is, is the wedge self-locking? 60 lb/ft P 8 B 15 ft
24 7.2 Wedges Example 7, page 2 of Impending motion of top of wedge relative to beam 3 P Free-body diagram of wedge Impending motion of top of wedge relative to floor y 8 f N N B f B 4 Equilibrium equations for wedge + Fx = 0: P f f B cos 8 N B sin = 0 (1) F y = 0: N + f B sin 8 N B cos = 0 (2) Slip impends so, f = f -max N = 0.3N (3) f B = f B-max N B = 0.3N B (4) 5 Geometry = 180 ( ) = = 82
25 7.2 Wedges Example 7, page 3 of 6 6 Impending motion of beam relative to wedge (n observer on the wedge would see the end of the beam move up and to the left.) f B 60 lb/ft B 7 The friction force from the wedge opposes the motion of the beam. 8 8 N B B y Bx 8 Equilibrium equation of the beam 15 ft M B = 0: (60 lb/ft)(15 ft)( 15 ft 2 ) N B cos 8 (15 ft) + f B sin 8 (15 ft) = 0 (5) Solving Eqs. 1-5 simultaneously gives f = 135 lb N = 450 lb f B = 142 lb N B = 474 lb P = 342 lb ns.
26 7.2 Wedges Example 7, page 4 of 6 9 Second part of the problem: Determine if the wedge is self-locking. To do so, reverse the direction of force P and calculate the value of P necessary to cause impending motion of the wedge to the left. 60 lb/ft P 8 B 15 ft
27 7.2 Wedges Example 7, page 5 of 6 10 Free-body diagram of wedge 11 Impending motion of top surface of wedge relative to beam (The force P has been applied to pull the wedge out.) y 8 N B 12 P Impending motion of bottom surface of wedge relative to floor 8 N f B f 13 + x Equilibrium equations for wedge Fx = 0: P + f + f B cos 8 N B sin 8 = 0 (1) F y = 0: N f B sin 8 N B cos 8 = 0 (2) Slip impends so, f = f -max N = 0.3N (3) f B = f B-max N B = 0.3N B (4)
28 7.2 Wedges Example 7, page 6 of 6 14 Free-body diagram of beam 60 lb/ft f B B N B 8 8 B y Bx 15 ft 17 Solving Eqs. 1-5 simultaneously gives 15 Impending motion of beam relative to wedge (an observer on the wedge sees the beam move down and to the right) 16 Equilibrium equation of the beam f = 135 lb N = 450 lb f B = 131 lb M B = 0: (60 lb/ft)(15 ft)( 15 ft 2 ) N B cos 8 (15 ft) f B sin 8 (15 ft) = 0 (5) N B = 436 lb, P = 204 lb. force P of at least 204 lb is necessary to pull out the wedge; any thing less than 204 lb is not enough the wedge will remain in place. In particular, when no force is appliced (P = 0), the wedge remains in place. Thus it is self-locking.
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