Tipos De Pavimentos

Problem 18.1 A horizontal force F = 133.4 N is applied to the 1023 N refrigerator as shown. Friction is negligible. (a) (b) What is the magnitude of the refrigerator’s acceleration? What normal forces are exerted on the refrigerator by the floor at A and B?
F

1524 mm 711.2 mm A ` 355.6 mm 355.6 mm

Solution: Assume that the refrigerator rolls without tipping. We have the following equationsof motion.
Fx : (133.4 N ) = 1023 N 9.81 m/s 2 a

F = 133.4 N 1023 N 1524 mm

Fy : A + B − 1 0 2 3 N = 0 MG : −(1 33 . 4 N )(0.813 m.) − A(0.356 m) + B(0 . 3 5 6 m ) = 0 Solving we find

711.2 mm
(a) (b) a = 1.28 m/s
2

A
A = 359 N , B = 664.1 N

B

Since A > 0 and B > 0 then our assumption is correct.

Problem 18.2 Solve Problem 18.1 if the coefficient of kinetic friction at A andB is µk = 0.1. Solution: Assume sliding without tipping
Fx : (133.4 N ) − (0.1)(A + B) = Fy : A + B − 1 0 2 3 N = 0 MG : −(1 3 3 . 4 N )(0.813 m ) − A(0.356 m ) + B(0.356 m ) − (0.1)(A + B)(0.711 m ) = 0 Solving, we find (a) (b) a = 0.3 m/s 2 A = 256.6 N , B = 765.1 N 1023 N 9.81 m/s
2

F = 133.4 N
a

1023 N 1524 mm

711.2 mm

µA
N A

µB
B

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469

Problem 18.3 As the 2800-N airplane begins its takeoff run at t = 0, its propeller exerts ahorizontal force T = 1000 N. Neglect horizontal forces exerted on the wheels by the runway. (a) What distance has the airplane moved at t = 2 s? (b) what normal forces are exerted on the tires at A and B? Solution: The unknowns are NA , NB , a.
The equations of motion are: Fx : −T = − W a, g
4m

T 3m A

W 5m B 2m

Fy : N A + N B − W = 0 MG : NB (2 m ) − N (5 m ) A + T (1 m ) = 0 Putting inthe numbers for T , W , and g and solving we find NA = 943 N, (a) NB = 1860 N, a = 3.5 m/s 2.

The distance is given by d = d =7 m

1 2 1 at = (3 .5 m/s 2)(2 s)2 = 7 m 2 2

(b)

The forces were found to be NA = 943 N, NB = 1860 N

Problem 18.4 The Boeing 747 begins its takeoff run at time t = 0. The normal forces exerted on its tires at A and B are NA = 175 kN and NB = 2800 kN. If youassume that these forces are constant and neglect horizontal forces other than the thrust T , how fast is the airplanes moving at t = 4 s? (See Active Example 18.1.)

T A 3m B 26 m 2m

5m

Solution: The unknowns are T , W, a. The equations of motion
are: Fx : −T = − W a, g

Fy : NA + NB − W = 0, MG : NB (2 m) − NA (24 m) − T (2 m) = 0. Putting in the numbers for NA and NB and solving, wefind a = 2.31 m/s2 , T = 700 kN, W = 2980 kN.

The velocity is then given by v = at = (2.31 m/s2 )(4 s) = 9.23 m/s. v = 9.23 m/s.

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Problem 18.5 The crane moves to the right with constant acceleration, and the 800-kg load moves without swinging. (a) (b) What is the acceleration of the crane and load? What are the tensions in the cables attached at A and B?
5° A 1m 5° B

1.5 m

1.5 m

Solution: From Newton’s second law: Fx = 800a N.
The sum ofthe forces on the load:

5˚ FA 1.0 m mg 1.5 m 1.5 m

5˚ FB

Fx = FA sin 5◦ + FB sin 5◦ − 800a = 0. Fy = FA cos 5◦ + FB cos 5◦ − 800g = 0. The sum of the moments about the center of mass: MCM = −1.5FA cos 5◦ + 1.5FB cos 5◦ − FA sin 5◦ − FB sin 5◦ = 0. Solve these three simultaneous equations: a = 0.858 m/s2 , FA = 3709 N , FB = 4169 N

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