By L. I. Sedov, J. R. M. Radok

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**Additional info for A course in continuum mechanics Vol. I Basic equations and analytical techniques **

**Sample text**

6 Distributed Loads Forces do not act necessarily at a single point, as has been demonstrated up to now. If we imagine, for instance, someone lying lazily stretched out on a plank bed (see Figure 19 a, b, c), the bed will be subjected to different loads along practically its entire length, depending on the anatomy of the person. Summing up the resultant of this distributed load generally requires mathematical integration. Here the distributed load q is specified as a function q(x) which describes a force per unit length (newtons per meter).

Knowing from above that L=2R, we can calculate the unknown force BX from the moment equation (III) and from equation II we immediately get AY: Bx= F 4/(3cosD) mg tanD, Ay = mg F sinD . By putting Bx into equation (I), we then get, after simple transformation: Ax = (cosD 4/(3cosD))F + mg tanD. ”). So when faced with such exercises or test questions, proceed as follows: After taking a deep breath, twist the corner of your mouth in a slightly contemptuous grin aimed in the direction of your nearest despairing classmate, then ask yourself the following two questions.

The first thing we do is draw a free-body diagram. At the start of every statics problem we therefore draw a correct free-body diagram. First we isolate the system by cutting it free. Cut it free, the only way to be. 53 Figure 28: Free-body diagram of the two warriors on ice. We therefore determine: 6 Fx = 0 = FR1 FR2 FR1 = FR2 P01 FN1 = P02 FN2. From the summation of forces in the y direction it then follows: P01 Mg = P02 (m + mR)g. This gives the mass of the backpack as: mR = (P01/P02)M m.