NOTE: - Refer to the given sample problem below on how to solve the problem.- The sub answers must be in 6 decimal places and the FINAL answer should be in 4 decimal places.- Every part should have a free body diagram to show how you solved the problem. - PLEASE FOLLOW MY INSTRUCTIONS. Thank you. A pulley-and-cable assembly supports block W and requires a downward force at P for equilibrium. Knowing that the tension of cable FG is 61.2 N,calculate the mass of the block and the magnitude of the downward force P.Aww54"L36*78.70125w.ww- Sample Problem 9The system shown is used tosupport a 10lb Block and aBlock A. If the Tension in Cable O2 is 5lb,following:a) WA weight of Block Ab) T, T3 tension at cables 1and 310-lbdetermine the T6045c) x spring deformation (k =30lb/ft)2. Equilibrium of ParticlesSample Problem 9Given:T2 = 5lbk = 30lb/ftA10 Ib60"45Required:a) WAb) T,, T3c) x deformation of sering2. Equilibrium of ParticlesSample Problem 9* Consider A31.10 lb6045-ST, cos 6o 1(5 )ws450T :: 357o lb. T3T, sınco + 54 5 sin a5-w-oW, 3(1570)sinwo 1545sinusW 14 6592 lo2. Equilibrium of ParticlesSample Problem 9* Consider the 10 l6 blockPrinciple ofTransomissibility10 lb6045-1, cos4s 1T301 Solve x-5 sinys -10 - Fg:0Fg= -15-535 N.TF: 13 535S N,CTyi35 395 1b, T-135355 : () *x-0 4512 4

Answered: Image /qna-images/answer/1b07b6c1-b38a-467f-a68a-5865345cea9f.jpg

A pulley-and-cable assembly supports block W and requires a downward force at P for equilibrium. Knowing that the tension of cable FG is 61.2 N, calculate the mass of the block and the magnitude of the downward force P.

A pulley-and-cable assembly supports block W and requires a downward force at P for equilibrium. Knowing that the tension of cable FG is 61.2 N, calculate the mass of the block and the magnitude of the downward force P.

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter4: Coplanar Equilibrium Analysis

Section: Chapter Questions

Problem 4.38P: The 40-kghomogeneous disk is placed on a frictionless inclined surface and held in equilibrium by...

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Concept explainers

Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.

Question

NOTE:

- Refer to the given sample problem below on how to solve the problem.

- The sub answers must be in 6 decimal places and the FINAL answer should be in 4 decimal places.

- Every part should have a free body diagram to show how you solved the problem.

- PLEASE FOLLOW MY INSTRUCTIONS. Thank you.

A pulley-and-cable assembly supports block W and requires a downward force at P for equilibrium. Knowing that the tension of cable FG is 61.2 N,
calculate the mass of the block and the magnitude of the downward force P.
A
ww
54"
L36*
78.701
25
w.
ww-

Sample Problem 9
The system shown is used to
support a 10lb Block and a
Block A. If the Tension in Cable O
2 is 5lb,
following:
a) WA weight of Block A
b) T, T3 tension at cables 1
and 3
10-lb
determine the T
60
45
c) x spring deformation (k =
30lb/ft)
2. Equilibrium of Particles
Sample Problem 9
Given:
T2 = 5lb
k = 30lb/ft
A10 Ib
60"
45
Required:
a) WA
b) T,, T3
c) x deformation of sering
2. Equilibrium of Particles
Sample Problem 9
* Consider A
31.
10 lb
60
45
-ST, cos 6o 1(5 )ws450
T :: 357o lb. T
3T, sınco + 54 5 sin a5-w-o
W, 3(1570)sinwo 1545sinus
W 14 6592 lo
2. Equilibrium of Particles
Sample Problem 9
* Consider the 10 l6 block
Principle of
Transomissibility
10 lb
60
45
-1, cos4s 1T30
1 Solve x
-5 sinys -10 - Fg:0
Fg= -15-535 N.T
F: 13 535S N,C
Tyi35 395 1b, T
-135355 : () *
x-0 4512 4

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