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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: abelo
Type : Application
Page(s) : 122
Taille Size: 30.68 Ko KB
Mis en ligne Uploaded: 03/12/2019 - 23:51:58
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PROBLEM 2.1

Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their
resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION
(a) Parallelogram law:

(b) Triangle rule:

We measure: R = 179 N, α = 75.1° R = 179 N 75.1° !

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

3
PROBLEM 2.3

The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.

SOLUTION

We measure: α = 51.3°
β = 59.0°
(a) Parallelogram law:

(b) Triangle rule:

We measure: R = 139.1 lb, γ = 67.0° R = 139.1 lb 67.0° !

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

5
PROBLEM 2.5

The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle α by trigonometry knowing that the component
along line a-a′ is to be 240 lb. (b) What is the corresponding value of the
component along b-b′?

SOLUTION
(a) Using the triangle rule and law of sines:
sin β sin 60°
=
240 lb 300 lb
sin β = 0.69282
β = 43.854°
α + β + 60° = 180°
α = 180° − 60° − 43.854°
= 76.146° α = 76.1°
Fbb′ 300 lb
(b) Law of sines: = Fbb′ = 336 lb
sin 76.146° sin 60°

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

7
PROBLEM 2.7

Two forces are applied as shown to a hook support. Knowing that the
magnitude of P is 35 N, determine by trigonometry (a) the required
angle α if the resultant R of the two forces applied to the support is to
be horizontal, (b) the corresponding magnitude of R.

SOLUTION
Using the triangle rule and law of sines:
sin α sin 25°
(a) =
50 N 35 N
sin α = 0.60374

α = 37.138° α = 37.1°
(b) α + β + 25° = 180°
β = 180° − 25° − 37.138°
= 117.86°
R 35 N
= R = 73.2 N
sin117.86 sin 25°

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

9
PROBLEM 2.9

A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that α = 25°, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical. (b) What is the corresponding magnitude of the
resultant?

SOLUTION

Using the triangle rule and the law of sines:
1600 N P
(a) = P = 3660 N
sin 25° sin 75°
(b) 25° + β + 75° = 180°
β = 180° − 25° − 75°
= 80°
1600 N R
= R = 3730 N !
sin 25° sin 80°

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

11
PROBLEM 2.11

A steel tank is to be positioned in an excavation. Knowing that
α = 20°, determine by trigonometry (a) the required magnitude
of the force P if the resultant R of the two forces applied at A is
to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines:
(a) β + 50° + 60° = 180°
β = 180° − 50° − 60°
= 70°
425 lb P
= P = 392 lb
sin 70° sin 60°

425 lb R
(b) = R = 346 lb !
sin 70° sin 50°

reproduced or distributed in any form or by any means, without the prior writ...

Archive contentsContenu de l'archive

Action(s) SizeTaille FileFichier
3.65 Ko KB lisezmoi.txt
1.09 Ko KB concreto.hpprgm
4.21 Ko KB concreto 001_020.hpappdir.zip
4.21 Ko KB concreto 021_040.hpappdir.zip
4.21 Ko KB concreto 041_060.hpappdir.zip
4.21 Ko KB concreto 061_080.hpappdir.zip
4.21 Ko KB concreto 081_100.hpappdir.zip
4.21 Ko KB concreto 101_120.hpappdir.zip
4.21 Ko KB concreto 121_122.hpappdir.zip
505 octets bytes appslist.txt

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