The force of static friction on the luggage is 300 N
Explanation:
In this situation, the luggage is not moving, so we are talking about static friction.
The force of static friction acts on the luggage preventing it from start moving, and it acts opposite to the force of 300 N applied on the other side, that tries to push the object.
The maximum force of static friction for an object on a flat surface is given by:
[tex]F_{max} = \mu_s mg[/tex]
where
[tex]\mu_s[/tex] is the coefficient of static friction
m is the mass of the object
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
In this problem,
m = 49 kg
[tex]\mu_s = 0.95[/tex]
So the value of the maximum force of friction is
[tex]F_{max}=(0.95)(49)(9.8)=456 N[/tex]
However, in this situation the force of friction is less than that. In fact, the object is pushed with a force of
F = 300 N
While the force of friction acting in the opposite direction is
[tex]F_f[/tex]
So the equation of motion would be
[tex]F-F_f = ma[/tex]
However, the luggage is at rest, so its acceleration is zero. Therefore:
a = 0
[tex]F-F_f = 0\rightarrow F_f = F = 300 N[/tex]
So, the force of friction is equal (and opposite) to the push applied, 300 N. In order to move the luggage, you have to apply a force F larger than the maximum force of friction, calculated before (456 N): below that value, the force of friction will "balance" the force applied, preventing the luggage from moving.
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