What is the coefficient of static friction between a load of mass 2 kg and the horizontal surface if the limited frictional force

The surface exerts an upward normal force N such that the net vertical force on the mass is

∑ F = N - mg = 0

where m = 2 kg and g is the acceleration due to gravity. So

N = (2 kg) (9.8 m/s²) = 19.6 N

The frictional force then has magnitude µ (19.6 N), where µ is the coefficient of static friction. If the maximum magnitude of static friction is to be 10 N, we have

10 N = µ (19.6 N)

===>   µ = (10 N)/(19.6 N) ≈ 0.51

0.33

Explanation:

R= 60N

coefficient of limiting friction = F/R

= 20/60 = 0.33

10

Explanation:

[tex]v\approx 16.956\,\frac{m}{s}[/tex]

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

[tex]\Sigma F = f = m\cdot \frac{v^{2}}{R}[/tex]

The maximum speed is:

[tex]v = \sqrt{\frac{f\cdot R}{m} }[/tex]

[tex]v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }[/tex]

[tex]v\approx 16.956\,\frac{m}{s}[/tex]

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The ratio between the force of the limiting friction and the normal reaction is coefficient of friction

idk what it lol

The maximum friction that can be generated between two static surfaces in contact with each other. Once a force applied to the two surfaces exceeds the limiting friction, motion will occur. For two dry surfaces, the limiting friction is a product of the normal reaction force and the coefficient of limiting friction

Explanation:

Hope it helps

when the moving force and the force opposing motion are equal; any addition to the moving force will cause slipping. The limiting frictional force is proportional to the normal reaction between the contacting surfaces and is independent of the area of contact.