*8-80. Determine the magnitude of the horizontal force P that must be applied to the handle of the bench vise in order to produce a clamping force of 600 N on the block. The single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is 4, = 0.25. 100 mm Here, M = P(0.1) 7.5 | 2x (12.5)] O, = tan - u, = tan ¬(0.25) = 14.036° e = tan 2 - 5.455° tan W = 600 N Thus M = Wr tan(6, +0) P(0.1) = 600(0.0125) tan(14.036° + 5.455°) P= 26.5 N Ans.

*8-80. Determine the magnitude of the horizontal force P that must be applied to the handle of the bench vise in order to produce a clamping force of 600 N on the block. The single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is 4, = 0.25. 100 mm Here, M = P(0.1) 7.5 | 2x (12.5)] O, = tan - u, = tan ¬(0.25) = 14.036° e = tan 2 - 5.455° tan W = 600 N Thus M = Wr tan(6, +0) P(0.1) = 600(0.0125) tan(14.036° + 5.455°) P= 26.5 N Ans.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Whats the free body diagram on this problem.

*8-80. Determine the magnitude of the horizontal force P
that must be applied to the handle of the bench vise in order
to produce a clamping force of 600 N on the block. The
single square-threaded screw has a mean diameter of
25 mm and a lead of 7.5 mm. The coefficient of static
friction is 4, = 0.25.
100 mm
Here, M = P(0.1)
7.5
| 2x (12.5)]
O, = tan - u, = tan ¬(0.25) = 14.036°
e = tan
2
- 5.455°
tan
W = 600 N
Thus
M = Wr tan(6, +0)
P(0.1) = 600(0.0125) tan(14.036° + 5.455°)
P= 26.5N
Ans.

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