Engineering Mechanics ENGG*1210, Winter 2013

Midterm Review

Assigned: Tuesday, February ,

-. e connected bar BC is used to increase the lever arm of the crescent wrench as shown.If a clockwise moment ofMA = N ·m is needed to tighten the nut at A and the extensionis d = mm, determine the required force F in order to develop this moment.

Midterm Review

-. e chain AB exerts a force of lb on the door at B. Determine the magnitude of themoment of this force along the hinged axis x of the door.

Midterm Review

-. A boy stands out at the end of the diving board, which is supported by two springs A andB, each having a stiffness of k = kN/m. In the position shown, the board is horizontal.If the boy has a mass of kg, determine the angle of tilt which the board makes with thehorizontal aer he jumps off. Neglect the weight of the board and assume it is rigid.

Midterm Review

-. Due to an unequal distribution of fuel in the wing tanks, the centres of gravity for theairplane fuselage A and wings B and C are located as shown. If these components haveweightsWA = lb,WB = lb, andWC = lb, determine the normal reactions ofthe wheels D, E, and F on the ground.

4–29.

The connected bar BC is used to increase the lever arm ofthe crescent wrench as shown. If a clockwise moment of

is needed to tighten the nut at A and theextension d = 300 mm, determine the required force F inorder to develop this moment.

SOLUTION

By resolving force F into components parallel and perpendicular to the box wrenchBC, Fig. a, the moment of F can be obtained by adding algebraically the momentsof these two components about point A in accordance with the principle ofmoments.

a

Ans.F = 239 N

+(MR)A = ©Fd; -120 = F sin 15°(0.3 sin 30°) - F cos 15°(0.3 cos 30° + 0.3)

MA = 120 N # m

300 mm

30�

15�d

C

B

A

F

4–50.

The chain AB exerts a force of 20 lb on the door at B.Determine the magnitude of the moment of this force alongthe hinged axis x of the door.

SOLUTIONPosition Vector and Force Vector:

Moment of Force F About x Axis: The unit vector along the x axis is i. Applying Eq. 4–11, we have

Ans

Ans.

.

Or

= 44.4 lb # ft

= 132.8191111.7122 - 1-11.102211.026124 - 0 + 0

= 3 1 0 00 2.8191 1.0261

11.814 -11.102 11.712

3Mx = i # 1rOB * F2

= 44.4 lb # ft

= 130111.7122 - 1-11.10221424 - 0 + 0

= 3 1 0 03 0 4

11.814 -11.102 11.712

3Mx = i # 1rOA * F2

= 511.814i - 11.102j + 11.712k6 lb

F = 20¢ 13 - 02i + 10 - 3 cos 20°2j + 14 - 3 sin 20°2k213 - 022 + 10 - 3 cos 20°22 + 14 - 3 sin 20°22 ≤ lb

= 52.8191j + 1.0261k6 ft

rOB = 510 - 02i + 13 cos 20° - 02j + 13 sin 20° - 02k6 ft

rOA = 513 - 02i + 14 - 02k6 ft = 53i + 4k6 ft

F

3 ft

3 ft

2 ft

4 ft

O B

y

x

z

20˚

= 20 lbA

*5–52.

SOLUTIONEquations of Equilibrium: The spring force at A and B can be obtained directly bysumming moments about points B and A, respectively.

a

a

Spring Formula: Applying , we have

Geometry: The angle of tilt is

Ans.a = tan- 1 a0.10464 + 0.078481

b = 10.4°

a

¢A =1177.215(103)

= 0.07848 m ¢B =1569.615(103)

= 0.10464 m

¢ =F

k

+ ©MA = 0; FB (1) - 392.4(4) = 0 FB = 1569.6 N

+ ©MB = 0; FA (1) - 392.4(3) = 0 FA = 1177.2 N

BA

1 m 3 m

A boy stands out at the end of the diving board, which is supported by two springs A and B, each having a stiffness of k = 15kN>m. In the position shown the board is horizontal. If the boy has a mass of 40 kg, determine the angle of tilt which the board makes with the horizontal after he jumps off. Neglect the weight of the board and assume it is rigid.

5–65.

8 ft20 ft

ABD

E

F

8 ft6 ft

6 ft

4 ft

3 ft

z

x

y

C

SOLUTION

Solving,

Ans.

Ans.

Ans.RF = 13.7 kip

RE = 22.6 kip

RD = 22.6 kip

©Fz = 0; RD + RE + RF - 8000 - 6000 - 45 000 = 0

©My = 0; 8000(4) + 45 000(7) + 6000(4) - RF (27) = 0

©Mx = 0; 8000(6) - RD (14) - 6000(8) + RE (14) = 0

Due to an unequal distribution of fuel in the wing tanks, thecenters of gravity for the airplane fuselage A and wings Band C are located as shown. If these components haveweights and determine the normal reactions of the wheels D, E, and Fon the ground.

WC = 6000 lb,WB = 8000 lb,WA = 45 000 lb,