Final answer:
The smallest and largest mass that can hang without moving the 20kg block on the table is approximately 5.4 kg, due to the maximum static frictional force being 52.92 N which is generated by the coefficient of static friction of 0.27. The strength of static friction is 52.92 N, and the direction is opposite the potential motion caused by the hanging mass.
Step-by-step explanation:
To find the smallest mass (m) that will keep the system at rest, we must use the coefficient of static friction (μs) which is 0.27. The static frictional force (fs) equals μs multiplied by the normal force (N). The normal force in this scenario is equal to the weight of the 20kg block, which is mg, where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s2), thus N = 20kg × 9.8 m/s2 = 196 N. The maximum static frictional force is fs = μs × N = 0.27 × 196 N = 52.92 N.
The smallest mass (m) to keep the system at rest must produce a tension that is equal to the maximum static frictional force (52.92 N). Since the tension in the string is equal to the weight of the hanging mass, m × g = fs, so the smallest mass m = fs / g = 52.92 N / 9.8 m/s2 ≈ 5.4 kg.
The strength of the friction in this scenario is 52.92 N, and the direction of the friction is opposite to the potential motion of the block, which would be towards the left side where the string is pulling.
To find the largest mass (m) that will keep the system at rest, we should consider the point at which the object is on the verge of moving due to the hanging mass. Since the maximum static frictional force has already been calculated, any additional mass would exceed this force and cause the block to move. Therefore, the largest mass m that can hang without moving the block is also 5.4 kg.
In this particular case, when the system is at rest, the strength of the friction remains 52.92 N, and the direction of the friction is unchanged, opposing the direction of the potential motion of the block.