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By G. H. Ryder, M. D. Bennett (auth.)

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Tan() 3 tan () 1-2tan 2 () in this case. It is clear that if 2 tan 2 () = 1 then P will be infinite, which means that it will not be possible to push the crate up the incline no matter how large P is made. This phenomenon of lock-up is often exploited in ratchets and other devices. 8 m above ground level. The car is at the bottom of a slope of 20°, facing upwards. 8. 17 shows a free-body diagram for the car attempting to drive up the hill with rear-wheel drive. The direction ofthe friction force F 2 on the driven rear axle is determined as follows.

15a, such as that carried out by Coulomb and others. The horizontal force P applied to the block is slowly increased from zero until it starts to slide. 15b. 15c will be obtained. For low values of P the block will not move and F will be equal to P, that is, the block will be in equilibrium. 15 Friction P 0 (c) 30 MECHANICS OF MACHINES and continue to slide with increasing speed as P is increased still further, but generally this is accompanied by a reduction in F. This reduction is quite sharp at first, but then tends to level off.

It is convenient to represent the forces at C and D as components along and perpendicular to CD. 12b. 5 cos 60°) = 0 hence F3=1317N Finally, taking moments about C for the member CD hence M= 1054 Nm Note that the solution has been obtained by moment equations only, two for the platform and one for the link CD. This was made possible by careful choice of the directions for the components of the force at D. Any other directions, say vertical and horizontal, would be equally valid, but might not result in quite such a simple solution.

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