The system in fig. 13-29 is in equilibrium, but it begins to slip if any additional mass is added to the 5.0 kg object. what is the coefficient of static friction between the 10 kg block and the plane on which it rests?

0.289

Explanation:

Since the body is in equilibrium ( no net movement), then the total upward force must equal the total downward force and also total forward force must equal total backward force

Tension in the rope inclined at 30⁰ to the vertical hhas vertical and horizontal components

vertical component = Tcos 30⁰ and horizontal component = T sin30⁰

T cos 30⁰ = the downward force = mg where m is the mass of 5kg and g is 9.8 m/s² acceleration due to gravity

T cos 30⁰ = 49

T in the inclined rope to the vertical = 49 / cos 30⁰ = 56.58 N

Horizontal component of the tension in the rope = T sin30⁰ = 56.58 sin 30⁰ = 28.29 N

since the 10kg was not moving the frictional force impeding the movement = the horizontal component of the T along the positive x axis since the frictional force will act along the negative x axis

Fr = 28.29 N

coefficient of static friction = Fr / normal (mg) = 28.29 / (10 × 9.8) = 0.289

a person with a mass of 106kg has a weight of 1060n

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

                                              ⇒      

here   k is the proportionality constant whose value is 1.

                                                            [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.