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Answer :
The coefficient of static friction of the given system is 0.0765.
Force required to get the tractor out of the mud = 176 lb, Weight of the tractor = 2300 lb;
We know that; The force required to move an object from a rest position is given by the product of the coefficient of static friction and the normal force acting on the object.
Mathematically, F = μsN where, F = force applied μs = coefficient of static friction, N = normal force acting on the object.
So, in the given situation, the force applied is 176 lb and the weight of the tractor is 2300 lb. The normal force acting on the tractor can be calculated by multiplying its weight by the acceleration due to gravity.
N = mg = 2300 x 32.2
= 7416 lb
Therefore, the coefficient of static friction can be calculated by;
μs = F / N
= 176 / 7416
= 0.0765
Thus, the coefficient of static friction of the given system is 0.0765.
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