What Is the Coefficient of Friction in an Equilibrium System?

[e4physics]

The person in the system above has a mass of 60 kg. If the system is in equilibrium, what is the coefficient of friction between the person and the ground, to two decimal places?

I need help on this question , I have done some work on it which is given below:

Converted masses into weights/forces:

for weight of man : 60*98.1 = 588.6 N

for weight of the hanging mass = 10 * 98.1 = 98.1 N

I have done the tension in the string , the man is pulling the weight.

T* COS60° = 98.1N

T = 98.1N/COS60°

T = 196.2 N

IF THE SYSTEM IS IN EQUILIBRIUM SUM OF THE FORCES SHOULD BE ZERO.

I HAVE ALSO FOUND THE OTHER COMPONENT OF T WHICH I THINK WILL BE EQUAL AND OPPOSITE THE FRICTIONAL FORCE. ITS GIVEN AS

Ty = 196.2* sin 60° = 169.9144 N

My understanding is that i need to use F = μ * W

But i am not sure about what will be W

Could someone explain this question and guide me if i have done this right.

 

[PhanthomJay]

You cannot determine the coef of static friction, but you can determine the min coef required to keep the mans feet from sliding, which perhaps is what the problem is asking? Your tension calc is wrong, based on a free body diagram of the hanging object, what is T?? Then look at the forces on the man, break up T into its horiz and vert components and solve for the friction force and normal force using equilibrium in x and y directions separately, then friction force = coef of friction times normal force at the
limiting value of static friction.

 

[e4physics]

Solution for friction equilibrium

THANKS FOR GUIDING ME - I NEED FURTHER HELP

I have solved this according to your advice. Confusion i have about angle. i did 90-60=30

x Component of T = 98.1* COS 30° = 84.95N
y component of T = 98.1 SIN 30° = 49.05


SUM X forces is only 84.95N (equal and opposite to friction force).

SUM of Y forces is = weight of man + y component of T
∑ Y forces = 588.6N + 49.05N = 637.65N

Friction Force = μ * normal reaction

μ = 84.95 N / 637.65N

μ = 0.13322

 

[PhanthomJay]

THANKS FOR GUIDING ME - I NEED FURTHER HELP

I have solved this according to your advice. Confusion i have about angle. i did 90-60=30

x Component of T = 98.1* COS 30° = 84.95N
y component of T = 98.1 SIN 30° = 49.05


SUM X forces is only 84.95N (equal and opposite to friction force).

.
Ok so far, good.

SUM of Y forces is = weight of man + y component of T
∑ Y forces = 588.6N + 49.05N = 637.65N

Friction Force = μ * normal reaction

μ = 84.95 N / 637.65N

μ = 0.13322

the weight of the man acts down on him. In which

direction does the y comp of T act on him? What about the normal force?

 

[e4physics]

the weight of the y component acts down on him and the y component of T acts upwards
so ∑ Y forces = 588.6N - 49.05N = 539.55

normal force will be equal in magnitude t the y foces

μ = 84.95 N / 539.55N = 0.1574
HAVE I UNDERSTOOD IT CORRECT LY??

 

[PhanthomJay]

the weight of the y component acts down on him and the y component of T acts upwards
so ∑ Y forces = 588.6N - 49.05N = 539.55

normal force will be equal in magnitude t the y foces

μ = 84.95 N / 539.55N = 0.1574



HAVE I UNDERSTOOD IT CORRECT LY??

YES, I think so, considering that the problem is not worded correctly or you copied it down wrong. The value of the static friction coefficient that you have calculated, which should be rounded to 0.16, is the minimum value required to prevent the man from sliding rightward. The actual value could be much greater than that, but the friction force would be the same 84.95 N.


-The Phantom