OpenStudy (anonymous):

2 years ago
OpenStudy (anonymous):

2 years ago
OpenStudy (anonymous):

number 1. coeff. 0.50

2 years ago
OpenStudy (anonymous):

@ganeshie8

2 years ago
OpenStudy (anonymous):

@e.mccormick

2 years ago
OpenStudy (anonymous):

@Preetha

2 years ago
OpenStudy (anonymous):

@nesha

2 years ago
OpenStudy (anonymous):

@Nixy

2 years ago
OpenStudy (anonymous):

@Nnesha

2 years ago
OpenStudy (anonymous):

@mathmate

2 years ago
OpenStudy (anonymous):

@e.mccormick

2 years ago
OpenStudy (anonymous):

|dw:1438058521329:dw|

2 years ago
OpenStudy (anonymous):

coeff. equal to 0.50

2 years ago
OpenStudy (aakashsudhakar):

I can help with the first question, regarding the block on the ledge at an unknown angle of inclination. Assuming the angle of inclination is just enough such that the block is in static equilibrium on the ledge, we can say the following force equilibrium expression is true. $\sum_{ }^{ }F_x= F_f \cos \theta - F_N \sin \theta = 0$ where Ff is the force of friction acting on the block in the direction of up the ledge, while FN is the normal force acting on the block perpendicular to the surface of the ledge. We can solve this out to obtain an expression of Ff, the force of friction, by itself. $F_f \cos \theta = F_N \sin \theta$ $F_f=F_N \tan \theta$ We already know the basic friction equation, given as follows: $F_f=\mu_f F_N$ Therefore, we can say the following is true: $\mu_f F_N = F_N \tan \theta$ Here we can see that we can easily factor out the normal force: $\mu_f=\tan \theta$ And now we finally have an equation to work with. We know the coefficient of friction to be equal to about 0.50, so we can quickly make the following expression: $\tan \theta = 0.50$ which rearranges to:$\theta = \tan^{-1}(0.50)$ Simply solve for that and you have your value for the angle of inclination.

2 years ago
OpenStudy (mathmate):

@jacalneaila When you tagged me yesterday, I came and did not see any new post, so I left. Sorry about that. Perhaps it would be more efficient if you could post the question before tagging for help. Also, @AakashSudhakar has done a super job of explaining the solution, so I gave him a medal probably at the expense of the previous receiver of the medal. Again, it would be nice if you could start a new post for each new problem so others can fairly give medals to all the helpful helpers. Thank you very much for you attention, and tag me any time. I like solving your problems.

2 years ago