Statics An 6q2t4d

PROBLEM 4.3 A T-shaped bracket s the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.

SOLUTION Free-Body Diagram:

ΣFx = 0: Bx = 0 ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0

(40a − 160) 12

A=

(1)

ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.) − (10 lb)(a + 20 in.) + (12 in.) B y = 0 By =

Bx = 0, B =

Since (a)

(b)

(1400 + 40a) 12 (1400 + 40a ) 12

(2)

For a = 10 in.,

Eq. (1):

A=

(40 × 10 − 160) = +20.0 lb 12

Eq. (2):

B=

(1400 + 40 × 10) = +150.0 lb 12

B = 150.0 lb W

Eq. (1):

A=

(40 × 7 − 160) = +10.00 lb 12

A = 10.00 lb W

Eq. (2):

B=

(1400 + 40 × 7) = +140.0 lb 12

B = 140.0 lb W

A = 20.0 lb W

For a = 7 in.,

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PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B:

Ty Tx

=

0.18 m 0.24 m

3 Ty = Tx 4

(a)

(1)

ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = +1600 N

From Eq. (1):

Ty =

3 (1600 N) = 1200 N 4

T = Tx2 + Ty2 = 16002 + 12002 = 2000 N

(b)

T = 2.00 kN W

ΣFx = 0: Cx − Tx = 0 Cx − 1600 N = 0 C x = +1600 N

C x = 1600 N

ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N

C y = 1680 N

α = 46.4° C = 2320 N

C = 2.32 kN

46.4° W

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PROBLEM 4.31 Neglecting friction, determine the tension in cable ABD and the reaction at C.

SOLUTION Free-Body Diagram:

ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0 ΣFx = 0: C x − 80 N = 0

C x = +80 N

ΣFy = 0: C y − 120 N + 80 N = 0

C y = +40 N

T = 80.0 N W C x = 80.0 N C y = 40.0 N C = 89.4 N

26.6° W

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PROBLEM 4.38 A light rod AD is ed by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C.

SOLUTION Free-Body Diagram:

ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0 A = 69.28 lb

A = 69.3 lb

W

ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30° + (69.28 lb)(8 in.)sin 30° = 0 C = 173.2 lb

C = 173.2 lb

60.0° W

B = 34.6 lb

60.0° W

ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30° + (69.28 lb)(16 in.) sin 30° = 0 B = 34.6 lb

Check:

ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0 0 = 0 (check)

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PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION Free-Body Diagram:

W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN

(a)

Rear wheels:

ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0

(3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0

A = +6.0659 kN

(b)

Front wheels:

A = 6.07 kN W

ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 3.4335 kN − 13.7340 kN + 2(6.0659 kN) + 2B = 0 B = +4.2346 kN

B = 4.23 kN W

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PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels.

SOLUTION Free-Body Diagram: W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα

From free-body diagram of hand truck, Dimensions in mm

ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0

(1)

ΣFy = 0: P − 2W + 2 B = 0

(2)

α = 35°

For

a1 = 300sin 35° − 80 cos 35° = 106.541 mm a2 = 430 cos 35° − 300sin 35° = 180.162 mm b = 930cos 35° = 761.81 mm

(a)

From Equation (1): P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0 P = 37.921 N

(b)

or P = 37.9 N W

From Equation (2): 37.921 N − 2(392.40 N) + 2 B = 0

or B = 373 N W

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PROBLEM 4.11 Three loads are applied as shown to a light beam ed by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P = 0.

SOLUTION

ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD

(1)

ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB

(2)

For cable B not to be slack, TB ≥ 0, and from Eq. (2), Q ≤ 11.00 kN

For cable D not to be slack, TD ≥ 0, and from Eq. (1), Q ≥ 0.500 kN

For neither cable to be slack, 0.500 kN ≤ Q ≤ 11.00 kN W

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PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.

SOLUTION Free-Body Diagram:

ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0 37.5 B= 0.25 − 0.866h

(a)

(1)

When h = 0, B=

From Eq. (1):

37.5 = 150 N 0.25

B = 150.0 N

30.0° W

ΣFy = 0: Ax − B sin 60° = 0 Ax = (150)sin 60° = 129.9 N

A x = 129.9 N

ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (150) cos 60° = 75 N

A y = 75 N

α = 30° A = 150.0 N

(b)

A = 150.0 N

30.0° W

When h = 200 mm = 0.2 m, From Eq. (1):

B=

37.5 = 488.3 N 0.25 − 0.866(0.2)

B = 488 N

30.0° W

ΣFx = 0: Ax − B sin 60° = 0 Ax = (488.3) sin 60° = 422.88 N

A x = 422.88 N

ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (488.3) cos 60° = −94.15 N

A y = 94.15 N

α = 12.55° A = 433.2 N

A = 433 N

12.55° W

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PROBLEM 4.35 A movable bracket is held at rest by a cable attached at C and by frictionless rollers at A and B. For the loading shown, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION Free-Body Diagram:

(a)

ΣFy = 0: T − 600 N = 0 T = 600 N W

(b)

ΣFx = 0: B − A = 0

∴ B=A

Note that the forces shown form two couples. ΣM = 0: (600 N)(600 mm) − A(90 mm) = 0 A = 4000 N ∴ B = 4000 N A = 4.00 kN

; B = 4.00 kN

W

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PROBLEM 4.46 A tension of 20 N is maintained in a tape as it es through the system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.

SOLUTION Free-Body Diagram:

ΣFx = 0: C x + (20 N) = 0

C x = −20 N

ΣFy = 0: C y − (20 N) = 0

C y = +20 N

C = 28.3 N

45.0° W

ΣM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0 M C = −4.30 N ⋅ m

M C = 4.30 N ⋅ m

W

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PROBLEM 4.54 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 N · m, P = 200 N, and l = 600 mm.

SOLUTION Free-Body Diagram:

(a)

From free-body diagram of rod AB: ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0 or sinθ + cosθ =

(b)

M W Pl

For M = 150 lb ⋅ in., P = 20 lb, and l = 6 in., sin θ + cos θ =

150 lb ⋅ in. 5 = = 1.25 (20 lb)(6 in.) 4 sin 2 θ + cos 2 θ = 1

Using identity

sin θ + (1 − sin 2 θ )1/2 = 1.25 (1 − sin 2 θ )1/2 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0

Using quadratic formula sin θ = =

or

−( −2.5) ± (625) − 4(2)(0.5625) 2(2) 2.5 ± 1.75 4

sin θ = 0.95572 and sin θ = 0.29428 θ = 72.886° and θ = 17.1144° or θ = 17.11° and θ = 72.9° W

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PROBLEM 4.56 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W = 75 lb, l = 30 in., and k = 3 lb/in.

SOLUTION Free-Body Diagram:

Spring force:

Fs = ks = k (l − l cos θ ) = kl (1 − cos θ )

§l · ΣM D = 0: Fs (l sin θ ) − W ¨ cos θ ¸ = 0 ©2 ¹

(a)

kl (1 − cos θ )l sin θ − kl (1 − cos θ ) tan θ −

(b)

For given values of

W l cos θ = 0 2

W =0 2

or (1 − cos θ ) tan θ =

W W 2kl

W = 75 lb l = 30 in. k = 3 lb/in. (1 − cos θ ) tan θ = tan θ − sin θ 75 lb = 2(3 lb/in.)(30 in.) = 0.41667

Solving numerically,

θ = 49.710°

or

θ = 49.7° W

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PROBLEM 4.58 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of θ corresponding to equilibrium.

SOLUTION First note:

T = ks

where

k = spring constant s = elongation of spring l = −l cos θ l (1 − cos θ ) = cos θ kl T= (1 − cos θ ) cos θ

(a)

From F.B.D. of collar B: or

(b)

For

ΣFy = 0: T sin θ − W = 0 kl (1 − cos θ )sin θ − W = 0 cos θ

or tan θ − sin θ =

W W kl

W = 3 lb l = 6 in. k = 8 lb/ft 6 in. l= = 0.5 ft 12 in./ft tan θ − sin θ =

Solving numerically,

3 lb = 0.75 (8 lb/ft)(0.5 ft)

θ = 57.957°

or

θ = 58.0° W

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