Required Information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. In the vise shown in the figure, the screw is single-threaded in the upper member; it passes through the lower member and is held by a frictionless washer. The pitch of the screw is 3 mm, its mean radius is 12 mm, and the coefficient of static friction is 0.15. D B M 450 mm 150 mm Determine the magnitude P of the forces exerted by the jaws when M= 74-N-m couple is applied to the screw. Assuming that the screw is single-threaded at both A and B (right-handed thread at A and left-handed thread at B). The value of magnitude Pis 24.22 kN. ΣMD: P (600 mm) - W (450 mm)=0 tan = 3 mm 2x(12 mm) P = 0.75 W Since screw is threaded in opposite senses at both A and B, the torque is T=20r P=0.75 ()cot 10.81° W = 0.75 ( 68 N.m 0.012 m x 2 -)cot 10.81° P 11.135 kN 0 = 2.28° stan 0.15 = 8.53° tan¹ = 0+, 0+ &s = 10.81° WQcot10.81° The free-body diagram of the upper member is

Structural Analysis
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Chapter2: Loads On Structures
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In the vise shown in the ffigure, the screw is single-threaded in the upper member: It passes through the lower member and is held by a frictionless washer. The pitch of the screw is 3 mm, its mean radius is 12 mm. and the coefficient of static friction is 0.15.

 

Determine the magnitude P of the forces exerted by the jaws when M= 74-N.m couple is applied to the screw. Assuming that the screw is single-threaded at both A and B (right-handed thread at A and left-handed thread at B)

 

The value of magnitude P is      KN?

 

Note: please consider everything. Refer to the image for more clarity. On occasions, I receive wrong answers!!. Kindly please show the magnitude of P which is the final answer in " KN" in the step by step working out please. Appreciate your time!. Side note, the answer in the image is wrong. This question has been very tricky!. And a lot of experts have given me wrong answers!. So I beg to show concise working out!. Also I've attached another image something to help with the question its a vague working out which I don't get so please implement that into the step by step working out. Thanks again, appreciate your time on this question!.

Required Information
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
In the vise shown in the figure, the screw is single-threaded in the upper member; it passes through the lower member
and is held by a frictionless washer. The pitch of the screw is 3 mm, its mean radius is 12 mm, and the coefficient of static
friction is 0.15.
D
B
M
450 mm
150 mm
Determine the magnitude P of the forces exerted by the jaws when M= 74-N-m couple is applied to the screw. Assuming that the
screw is single-threaded at both A and B (right-handed thread at A and left-handed thread at B).
The value of magnitude Pis 24.22 kN.
Transcribed Image Text:Required Information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. In the vise shown in the figure, the screw is single-threaded in the upper member; it passes through the lower member and is held by a frictionless washer. The pitch of the screw is 3 mm, its mean radius is 12 mm, and the coefficient of static friction is 0.15. D B M 450 mm 150 mm Determine the magnitude P of the forces exerted by the jaws when M= 74-N-m couple is applied to the screw. Assuming that the screw is single-threaded at both A and B (right-handed thread at A and left-handed thread at B). The value of magnitude Pis 24.22 kN.
ΣMD: P (600 mm)
-
W (450 mm)=0
tan =
3 mm
2x(12 mm)
P = 0.75 W
Since screw is threaded in opposite senses at both A and B, the torque is
T=20r
P=0.75 ()cot 10.81°
W = 0.75 (
68 N.m
0.012 m x 2
-)cot 10.81°
P 11.135 kN
0 = 2.28°
stan 0.15 = 8.53°
tan¹
=
0+,
0+ &s
= 10.81°
WQcot10.81°
The free-body diagram of the upper member is
Transcribed Image Text:ΣMD: P (600 mm) - W (450 mm)=0 tan = 3 mm 2x(12 mm) P = 0.75 W Since screw is threaded in opposite senses at both A and B, the torque is T=20r P=0.75 ()cot 10.81° W = 0.75 ( 68 N.m 0.012 m x 2 -)cot 10.81° P 11.135 kN 0 = 2.28° stan 0.15 = 8.53° tan¹ = 0+, 0+ &s = 10.81° WQcot10.81° The free-body diagram of the upper member is
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