I have done this on my own, but i need some clarificationbefore i hand it in to

Question

I have done this on my own, but i need some clarificationbefore i hand it in to the professor. Your help isappreciated. I need to model this situation... A person standing to the right exerts a 620 N force on a 70 kgblock at a 40 degree angle above the horizontal. The coefficient offriction is (0.5, 0.4) I have done this on my own, but i need some clarificationbefore i hand it in to the professor. Your help isappreciated. I need to model this situation... A person standing to the right exerts a 620 N force on a 70 kgblock at a 40 degree angle above the horizontal. The coefficient offriction is (0.5, 0.4)

Explanation / Answer

              Given Force   F = 620 N                         mass of the block   m = 70 kg                          angle           = 40 deg                        The coefficient of friction      =?                              F = mg cos                               = F sin / mg cos                                  = 620 sin 40/ 70 x 9.8 x cos 40                                   = 0.75

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