Composition and Resolution ForcesProblem 1-1:Determine a pair of horizontal and vertical components of the340-lb force of Fig. P1-1.340bAns.15Fx = 1600 Ib →8.Fy= 3000 Ib ↑Problem 1-2:Determine a set of horizontal and vertical components of the200b200-lb force of Fig. P1-2.Ans.Fx= 120 Ib -Fy =160 Ib 4Problem 1-3:Resolve the 100-lb force of Fig. P1-3 into horizontal and verticalcomponents for each of the following values of: (a) 0 = 22° (b) 0 = 78(С) 0 — 132°Ans.(a) Fx = 37.46 Ib →Fy= 92.718 Ib(b) Fx = 97.81 Ib →Fy = 20.79 Ib(b) Fx = 74.31 Ib -Fy = 66.9 Ib ↑%3D100bProblem 1-4:Resolve the 500-lb force of Fig. P1-4 into two components: ashearing component parallel to AB and a normal component perpendicular toАВ.500 ibAns.FH= 400 IbFy= 300 Ib13'

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Problem 1-1:Determine a pair of horizontal and vertical components of the 340-lb force of Fig. P1-1. 340b

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.4.9P: Two steel wines support a moveable overhead camera weighing W = 28 lb (see figure part a) used For...

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Concept explainers

Types of loading

In mechanical terminology, the term load indicates that external force/weight, either constant or variable, acts on a mechanical member/structure. The weight of an object of a specific material is also an external load, and the external force that works on a structure comes under the categories of loading. The external load generates its effects as stress and strain in the structure.

Question

Composition and Resolution Forces
Problem 1-1:Determine a pair of horizontal and vertical components of the
340-lb force of Fig. P1-1.
340b
Ans.
15
Fx = 1600 Ib →
8.
Fy
= 3000 Ib ↑
Problem 1-2:Determine a set of horizontal and vertical components of the
200b
200-lb force of Fig. P1-2.
Ans.
Fx
= 120 Ib -
Fy =
160 Ib 4
Problem 1-3:Resolve the 100-lb force of Fig. P1-3 into horizontal and vertical
components for each of the following values of: (a) 0 = 22° (b) 0 = 78
(С) 0 — 132°
Ans.
(a) Fx = 37.46 Ib →
Fy
= 92.718 Ib
(b) Fx = 97.81 Ib →
Fy = 20.79 Ib
(b) Fx = 74.31 Ib -
Fy = 66.9 Ib ↑
%3D
100b
Problem 1-4:Resolve the 500-lb force of Fig. P1-4 into two components: a
shearing component parallel to AB and a normal component perpendicular to
АВ.
500 ib
Ans.
FH
= 400 Ib
Fy
= 300 Ib
1
3'

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