Two people are playing tug-of-war. Due to their choice of footwear, theircoefficient of static friction is different. Participant 1 has a mass of 60 kg, acoefficient of static friction of 2.0, and can pull with a maximum force of1000 N. Participant 2 has a mass of 80 kg and a coefficient of staticfriction of 1.2, and can pull with a maximum force of 1200 N. Who wins?

Participant 1 wins

Explanation:

Coefficient of static, μ = F/N, where F is the frictional force and N is the normal force.

The force exerted by each participant is the sum of the frictional force acting on each of them and the maximum force with which each participant pulls on the rope.

Frictional force, F =  μ * N

Normal force, N = mass * acceleration due to gravity, g

For Participant 1; μ = 2.0, mass = 60 kg, g = 9.8 m/s²

Frictional force = 2.0 * 60 * 9.8 = 1176 N

Total force = (1176 + 1000) = 2176 N

For Participant 2; μ = 1.2, mass = 80 kg, g = 9.8 m/s²

Frictional force = 1.2 * 80 * 9.8 = 940.8N

Total force = (940.8 + 1200) N = 2140.8 N

From the values obtained above, Participant 1 exerts more force than Participant 2, therefore, Participant 1 wins