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This is a problem from the book Engineering Mechanics Statics vol-1 by Meriam Kraige

The rectangular steel yoke is used to prevent slippage between the two boards under tensile loads $P$ .If the coefficients of static friction between the yoke and the board surfaces and between the boards are all $0.30$ the question is to determine the maximum value of $h$ for which there is no slipping and to find out the corresponding normal force $N$ between the two boards for $P=800N$ if impending motion occurs at all surfaces. enter image description here

Here is what I tried

For the first part if $\theta$ denotes the angle between the top part of yoke and horizontal then $\cos \theta = \frac{200}{210}=\frac{150}{h}$ which gives $h=157.5 mm$

For the second part I tried to draw the FBD of the top block.

enter image description here

From FBD $$P=f_1+f_2+f_3+f_4$$ and $$mg=N_3-N_1$$ Also by taking moments about COM I got $P=f_3-f_1$.However I am not able to proceed after this.Any help shall be highly appreciated.Thanks

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  • $\begingroup$ To my understanding, the second part is quite direct: you have the force P, the coefficient and you need the normal force N. For the first part, I have not a clear idea, but your numbers seem not correct: you use the horizontal values which have no impact and don't use the important value of the coefficient. $\endgroup$ – Adrian Maire Sep 23 '17 at 21:52
  • $\begingroup$ @AdrianMarie I couldn't get your last line.can you please elaborate a bit. $\endgroup$ – user471651 Sep 24 '17 at 3:20
  • $\begingroup$ You don't need the $N_2$ and $N_3$ because the yoke is wider than the board, therefore, it can be assumed that there is no contact between them. $\endgroup$ – Jem Eripol Sep 25 '17 at 0:42
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I will try to answer it.

First, draw the forces acting upon it. (I considered only the top plank/board as it is symmetric with the bottom board). Refer to the figure below.

Reference drawing

Separating the upper plank and analyzing the forces upon it, you will get the figure below.

Figure 2

Note that $F_{\text f1}=P_y (0.3)$ and $F_{\text f2}=N(0.3)$, therefore:

Summing up forces in the x-axis gives:

$$0.3(P_y)+0.3(N)=800$$

and Summing up forces in the y-axis gives:

$$N-P_y=0$$

Solving both,

$$P_y=N=1,333.33\ N$$

Solving also for $F_{\text f1}$ gives $F_{\text f1}=400\ N.$

Solving for the angle of inclination of the yoke gives

$$\theta = arctan (400/1333.33)=16.70\ degrees$$

therefore, $h = 150/\cos(16.70)=156.60\ mm$

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  • $\begingroup$ There is a shortcut to this. :) $\endgroup$ – Jem Eripol Sep 25 '17 at 3:44

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