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

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)
=Fcos40/Fsin40+mg=620cos40/620sin40+70*9.8=0.4379
hence the coefficient of friction=0.4

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---