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•13–1. The casting has a mass of 3 Mg. Suspended in a vertical position and initially at rest, it is given an upward speed of 200 mm s in 0.3 s using a crane hook H. Determine the tension in cables AC and AB during this time interval if the acceleration is constant.
Kinematics: Applying the equation , we have
Equations of Motion:
F (^) AB = FAC = F = 18146.1 N = 18.1 kN Ans.
- c ©Fy = may ; 2 Fcos 30° - 29430 = 3000(0.6667)
FAB = FAC = F
:+ ©Fx = max; FAB sin 30 - FAC sin 30° = 0
(+ c) 0.2 = 0 + a(0.3) a = 0.6667 m>s^2
y = y 0 + ac t
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30 30
B C
A
H
13–2. The 160-Mg train travels with a speed of when it starts to climb the slope. If the engine exerts a traction force F of of the weight of the train and the rolling resistance is equal to of the weight of the train, determine the deceleration of the train.
F D 1 > 500
80 km>h
Free-Body Diagram: The tractive force and rolling resistance indicated on the free-
body diagram of the train, Fig. (a), are and
, respectively.
Equations of Motion: Here, the acceleration a of the train will be assumed to be directed up the slope. By referring to Fig. (a),
a = -0.5057 m>s^2 Ans.
+Q©Fx¿ = max¿ ; 78 480 - 3139.2 - 160(10^3 )(9.81)¢
≤ = 160(10^3 )a
F (^) D = a
b(160)(10^3 )(9.81)N = 3139.2 N
F = a
b(160)(10^3 )(9.81) N = 78 480 N
F
10
1
13–3. The 160-Mg train starts from rest and begins to climb the slope as shown. If the engine exerts a traction force F of of the weight of the train, determine the speed of the train when it has traveled up the slope a distance of 1 km. Neglect rolling resistance.
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F
10
1
Free-Body Diagram: Here, the tractive force indicated on the free-body diagram of
the train, Fig. (a), is.
Equations of Motion: Here, the acceleration a of the train will be assumed directed up the slope. By referring to Fig. (a),
Kinematics: Using the result of a ,
v = 22.4 m>s Ans.
v^2 = 0 + 2(0.2501)(1000 - 0)
A +QB v^2 =^ v 0 2 +^2 ac(s^ -^ s 0 )
a = 0.2501 m>s^2
+Q©Fx¿ = max¿ ; 196.2(10^3 ) - 160(10^3 )(9.81)a
b = 160(10^3 )a
F =
(160)(10^3 )(9.81) N = 196.2(10^3 ) N
*13–4. The 2-Mg truck is traveling at 15 m s when the brakes on all its wheels are applied, causing it to skid for a distance of 10 m before coming to rest. Determine the constant horizontal force developed in the coupling C , and the frictional force developed between the tires of the truck and the road during this time.The total mass of the boat and trailer is 1 Mg.
Kinematics: Since the motion of the truck and trailer is known, their common acceleration a will be determined first.
Free-Body Diagram: The free-body diagram of the truck and trailer are shown in Figs. (a) and (b), respectively. Here, F representes the frictional force developed when the truck skids, while the force developed in coupling C is represented by T.
Equations of Motion: Using the result of a and referrning to Fig. (a),
Ans.
Using the results of a and T and referring to Fig. (b),
F = 33 750 N = 33.75 kN Ans.
- c ©Fx = max ; 11 250 - F = 2000(-11.25)
T = 11 250 N = 11.25 kN
:+ ©Fx = max ; - T = 1000(-11.25)
a = -11.25 m>s^2 = 11.25 m>s^2 ;
0 = 15 2 + 2 a(10 - 0)
a :+^ b v^2 = v 0 2 + 2 ac(s - s 0 )
C
13–6. Motors A and B draw in the cable with the accelerations shown. Determine the acceleration of the 300-lb crate C and the tension developed in the cable. Neglect the mass of all the pulleys.
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B^ A P ¿ P
C aP ¿ 2 ft/s^2 aP 3 ft/s^2
Kinematics: We can express the length of the cable in terms of , , and by referring to Fig. (a).
The second time derivative of the above equation gives
(1)
Here, and. Substituting these values into Eq. (1),
Ans.
Free-Body Diagram: The free-body diagram of the crate is shown in Fig. (b).
Equations of Motion: Using the result of a C and referring to Fig. (b),
T = 162 lb Ans.
- c ©Fy = may ; 2 T - 300 =
aC = -2.5 ft>s^2 = 2.5 ft>s^2 c
3 + 2 + 2 aC = 0
aP = 3 ft>s^2 aP¿ = 2 ft>s^2
A + T^ B aP +^ aP¿ +^2 aC =^0
sP + sP¿ + 2 sC = l
sPsP¿ sC
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13–7. The van is traveling at 20 km h when the coupling of the trailer at A fails. If the trailer has a mass of 250 kg and coasts 45 m before coming to rest, determine the constant horizontal force F created by rolling friction which causes the trailer to stop.
:+ ©Fx = max ; F = 250(0.3429) = 85.7 N Ans.
a = -0.3429 m>s^2 = 0.3429 m>s^2 :
0 = 5.556 2 + 2(a)(45 - 0)
a ;+^ b y^2 = y^20 + 2 ac (s - s 0 )
20 km>h =
20(10^3 )
= 5.556 m>s
A
20 km/h
F
•13–9. Each of the three barges has a mass of 30 Mg, whereas the tugboat has a mass of 12 Mg. As the barges are being pulled forward with a constant velocity of 4 m s, the tugboat must overcome the frictional resistance of the water, which is 2 kN for each barge and 1.5 kN for the tugboat. If the cable between A and B breaks, determine the acceleration of the tugboat.
A B
1.5 kN
4 m/s
2 kN 2 kN 2 kN
Equations of Motion: When the tugboat and barges are travelling at a constant velocity, the driving force F can be determined by applying Eq. 13–7.
If the cable between barge A and B breaks and the driving force F remains the same, the acceleration of the tugboat and barge is given by
a = 0.0278 m>s^2 Ans.
:+ ©Fx = max ; (7.50 - 1.5 - 2 - 2) A 103 B = (12 000 + 30 000 + 3000)a
:+^ ©Fx = max ; F - 1.5 - 2 - 2 = 0 F = 7.50 kN
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13–10. The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. If the magnitude of P is increased until the crate begins to slide, determine the crate’s initial acceleration if the coefficient of static friction is and the coefficient of kinetic friction is mk = 0.3.
ms = 0.
Equations of Equilibrium: If the crate is on the verge of slipping,. From FBD(a),
(1)
(2)
Solving Eqs.(1) and (2) yields
Equations of Motion: The friction force developed between the crate and its contacting surface is since the crate is moving. From FBD(b),
a = 1.66 m>s^2 Ans.
:+ ©Fx = max ; 353.29 cos 20° - 0.3(663.97) = 80 a
N = 663.97 N
- c ©Fy = may ; N - 80(9.81) + 353.29 sin 20° = 80(0)
Ff = mkN = 0.3N
P = 353.29 N N = 663.97 N
:+ ©Fx = 0; P cos 20° - 0.5N = 0
- c ©Fy = 0; N + P sin 20° - 80(9.81) = 0
Ff = ms N = 0.5N
20
p
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13–11. The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. Determine the crate’s acceleration in if the coefficient of static friction is the coefficient of kinetic friction is and the towing force is P = (90t^2 ) N, where t is in seconds.
mk = 0.3,
ms = 0.4,
t = 2 s
Equations of Equilibrium: At ,. From FBD(a)
Since , the crate accelerates.
Equations of Motion: The friction force developed between the crate and its contacting surface is since the crate is moving. From FBD(b),
a = 1.75 m>s 2 Ans.
:+ ©Fx = max ; 360 cos 20° - 0.3(661.67) = 80 a
N = 661.67 N
- c ©Fy = may ; N - 80(9.81) + 360 sin 20° = 80(0)
Ff = mkN = 0.3N
F (^) f 7 (Ff)max = ms N = 0.4(661.67) = 264.67 N
:+ ©Fx = 0; 360 cos 20° - Ff = 0 Ff = 338.29N
- c ©Fy = 0; N + 360 sin 20° - 80(9.81) = 0 N = 661.67 N
t = 2 s P = 90 A 22 B = 360 N
20
p
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•13–13. The two boxcars A and B have a weight of 20 000 lb and 30 000 lb, respectively. If they coast freely down the incline when the brakes are applied to all the wheels of car A causing it to skid, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is. The wheels of car B are free to roll. Neglect their mass in the calculation. Suggestion: Solve the problem by representing single resultant normal forces acting on A and B , respectively.
mk = 0.
Car A :
Both cars:
Solving,
T = 5.98 kip Ans.
a = 3.61 ft>s 2
+Q©Fx = max ; 0.5(19 923.89) - 50 000 sin 5° = a
ba
+Q©Fx = max ; 0.5(19 923.89) - T - 20 000 sin 5° = a
ba
- a©Fy =^ 0;^ NA -^ 20 000 cos 5°^ =^0 NA =^ 19 923.89 lb
C
5
A B
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13–14. The 3.5-Mg engine is suspended from a spreader beam AB having a negligible mass and is hoisted by a crane which gives it an acceleration of when it has a velocity of 2 m s. Determine the force in chains CA and CB during the lift.
4 m>s^2
System:
Joint C :
T = TCA = TCB = 27.9 kN Ans.
- c ©Fy = may ; 48.335 - 2 T cos 30° = 0
T¿ = 48.335 kN
+ c ©Fy = may ; T¿ - 3.5 A 103 B(9.81) = 3.5 A 103 B (4)
60
C
B
D E
A
60
13–15. The 3.5-Mg engine is suspended from a spreader beam having a negligible mass and is hoisted by a crane which exerts a force of 40 kN on the hoisting cable. Determine the distance the engine is hoisted in 4 s, starting from rest.
System:
s = 0 + 0 + Ans.
(1.619)(4)^2 = 12.9 m
A + c^ B s = s 0 + y 0 t +
ac t^2
a = 1.619 m>s^2
+ c ©Fy = may ; 40 A 103 B - 3.5A 103 B(9.81) = 3.5A 103 Ba
60
C
B
D E
A
60
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•13–17. A force of is applied to the cord. Determine how high the 30-lb block A rises in 2 s starting from rest. Neglect the weight of the pulleys and cord.
F = 15 lb
Block:
s = 64.4 ft Ans.
s = 0 + 0 +
(32.2)(2)^2
(+ c) s = s 0 + y 0 t +
ac t^2
aA = 32.2 ft>s^2
- c ©Fy = may ; - 30 + 60 = a
baA
F A
B
C
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13–18. Determine the constant force F which must be applied to the cord in order to cause the 30-lb block A to have a speed of 12 ft/s when it has been displaced 3 ft upward starting from rest. Neglect the weight of the pulleys and cord.
F = 13.1 lb Ans.
- c ©Fy = may ; - 30 + 4 F = a
b(24)
a = 24 ft>s^2
(12)^2 = 0 + 2(a)(3)
A + c^ B y^2 = y^20 + 2 ac (s - s 0 )
F A
B
C
*13–20. The 10-lb block A travels to the right at at the instant shown. If the coefficient of kinetic friction is between the surface and A , determine the velocity of A when it has moved 4 ft. Block B has a weight of 20 lb.
mk = 0.
vA = 2 ft>s
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Block A :
Weight B :
Kinematics:
Solving Eqs. (1)–(3):
y = 11.9 ft>s Ans.
y^2 = (2)^2 + 2(17.173)(4 - 0)
y^2 = y^20 + 2 ac (s - s 0 )
aA = -17.173 ft>s^2 aB = 8.587 ft>s^2 T = 7.33 lb
aA = - 2 aB
sA + 2 sB = l
- T ©Fy = may ; 20 - 2 T = a
baB
;+ ©Fx = max ; - T + 2 = a
baA
A
B
•13–21. Block B has a mass m and is released from rest when it is on top of cart A , which has a mass of 3 m. Determine the tension in cord CD needed to hold the cart from moving while B slides down A. Neglect friction.
Block B :
Cart:
T = a Ans.
mg 2
bsin 2u
T = mg sin u cos u
:+ ©Fx = max ; - T + NB sin u = 0
NB = mg cos u
+a©Fy = may ; NB - mg cos u = 0
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A u
B
D C
13–22. Block B has a mass m and is released from rest when it is on top of cart A , which has a mass of 3 m. Determine the tension in cord CD needed to hold the cart from moving while B slides down A. The coefficient of kinetic friction between A and B is mk.
Block B :
Cart:
T = mg cos u(sin u - mk cos u) Ans.
:+ ©Fx = max ; - T + NB sin u - mk NB cos u = 0
NB = mg cos u
+a©Fy = may ; NB - mg cos u = 0
A u
B
D C
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*13–24. If the force of the motor M on the cable is shown in the graph, determine the velocity of the cart when The load and cart have a mass of 200 kg and the car starts from rest.
t = 3 s.
Free-Body Diagram: The free-body diagram of the rail car is shown in Fig. (a).
Equations of Motion: For ,. By referring to Fig. (a), we can write
For ,. Thus,
Equilibrium: For the rail car to move, force 3F must overcome the weight component of the rail crate. Thus, the time required to move the rail car is given by
Kinematics: The velocity of the rail car can be obtained by integrating the kinematic equation,. For , at will be used as the integration limit. Thus,
When ,
y = 1.125(3)^2 - 4.905(3) + 5.34645 = 0.756m>s Ans.
t = 3 s
= A1.125t^2 - 4.905t + 5.34645Bm>s
y = A1.125t^2 - 4.905tB 2
t
2.18 s
L
y
0
dy = L
t
2.18 s
(2.25t - 4.905)dt
A + c^ B
L
dy = L
adt
d v = adt 2.18 s … t 6 3 s v = 0 t = 2.18 s
©Fx¿ = 0; 3(150t) - 200(9.81) sin 30° = 0 t = 2.18 s
a = 1.845 m>s^2
+Q©Fx¿ = max¿ ; 3(450) - 200(9.81) sin 30° = 200 a
t 7 3 sF = 450 N
a = (2.25t - 4.905) m>s^2
+Q©Fx¿ = max¿ ; 3(150t) - 200(9.81) sin 30° = 200 a
F =
0 … t 6 3 s t = A 150 tB N
C
M F
F (N)
450
3
t (s)
30
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•13–25. If the motor draws in the cable with an acceleration of , determine the reactions at the supports A and B. The beam has a uniform mass of 30 kg m, and the crate has a mass of 200 kg. Neglect the mass of the motor and pulleys.
3 m>s^2
However
y = 22 gh Q.E.D.
y^2 2
= gh
L
y
0
y dy = L
h
0
g dy
y dy = at ds = g sin u ds dy = ds sin u
+R©Ft = mat ; mg sin u = mat at = g sin u
C
A B
2.5 m 3 m
0.5 m
3 m/s^2
13–26. A freight elevator, including its load, has a mass of 500 kg. It is prevented from rotating by the track and wheels mounted along its sides. When , the motor M draws in the cable with a speed of 6 m s, measured relative to the elevator. If it starts from rest, determine the constant acceleration of the elevator and the tension in the cable. Neglect the mass of the pulleys, motor, and cables.
t = 2 s
Ans.
T = 1320 N = 1.32 kN Ans.
- c ©Fy = may ; 4 T - 500(9.81) = 500(0.75)
aE = 0.75 m>s^2 c
1.5 = 0 + aE (2)
A + c^ B y = y 0 + ac t
yE = -
= -1.5 m>s = 1.5 m>s c
A + T^ B yP =^ yE +^ yP>E
3 yE = -yP
3 sE + sP = l
M
