Question #45620

Determine the appropriate coefficient of friction in each case. show all you solutions.

1.It takes 59 N of horizontal force to get a 22 kg leather suitcase just starting to move across a floor. find us (hint fn=fg=mg)

2. A horizontal force of 54 N keeps the suitcase in moving at a constant velocity. find uk (hint fn=fg=mg)

Expert's answer

Answer on Question #45620 – Physics, Mechanics | Kinematics | Dynamics

Determine the appropriate coefficient of friction in each case. show all you solutions.

1. It takes 59N59\,N of horizontal force to get a 22kg22\,kg leather suitcase just starting to move across a floor. find us (hint fn=fg=mgfn=fg=mg)

2. A horizontal force of 54N54\,N keeps the suitcase in moving at a constant velocity. find uk (hint fn=fg=mgfn=fg=mg)

By the definition of friction:


F=μNF = \mu N


Where μ\mu – is a coefficient of friction, NN – normal force.

From this equation we can find coefficient of friction:


μ=FN\mu = \frac{F}{N}


On a horizontal plane, normal force will be equal to the weight of the body.


μ=Fmg\mu = \frac{F}{mg}


1. In the first case we have static friction, so coefficient of static friction:


μst=Fstmg\mu_{st} = \frac{F_{st}}{mg}μst=59N22kg9.8ms20.27\mu_{st} = \frac{59\,N}{22\,kg \cdot 9.8\, \frac{m}{s^2}} \approx 0.27


2. When suitcase is moving with constant velocity, we will get kinetic friction:


μk=Fkmg\mu_{k} = \frac{F_{k}}{mg}μk=54N22kg9.8ms20.25\mu_{k} = \frac{54\,N}{22\,kg \cdot 9.8\, \frac{m}{s^2}} \approx 0.25


Answer: Coefficients of friction are:

- static friction μst0.27\mu_{st} \approx 0.27

- kinetic friction μk0.25\mu_{k} \approx 0.25

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