11. We denote the magnitude of 110 N force exerted by the worker on the crate as F. The magnitude of the static frictional force can vary between zero and ,maxssNfFμ=.(a) In this case, application of Newton’s second law in the vertical direction yields NFmg=. Thus, ()()220.37 35kg (9.8m /s ) 1.3 10 NssNsfFmgμμ====×which is greater than F.(b) The block, which is initially at rest, stays at rest since F< fs, max. Thus, it does not move.(c) By applying Newton’s second law to the horizontal direction, that the magnitude of the frictional force exerted on the crate is 21.1 10 Nsf=×.(d) Denoting the upward force exerted by the second worker as F2, then application of Newton’s second law in the vertical direction yields
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.