Question

A Tyrannosaurus is pulling on the body of a freshly killed, 6,000 kg Ankylosaurus with a force of 25,000N to the North. A second Tyrannosaurus is pulling to the East with a force of 15,000N…

a. What force of static friction must be present between the Ankylosaurus and the ground to make the net force zero?

b. What is the coefficient of friction between the Ankylosaurus armored skin and the ground?

Answer 1

1) In order to define the force of static friction, one has to calculate the net pulling force created by both Tyrannosaurs. As long as they pull in the perpendicular directions (N and E), the net pulling force can be obtained by means of the Pythagorean theorem:

$$F_{pull} = \sqrt{F_1^2 + F_2^2} = \sqrt{25^2 + 15^2} \cdot 10^3 \, N \approx \, 29155 \, N$$

The net force will be equal to zero, if the force of static friction is equal to the net pulling force in magnitude and opposite in direction. Hence,

$$F_{s.f} = F_{pull} = \, 29155 \, N$$

2) the coefficient of friction is

$$\mu = \frac{F_{s.f.}}{N} = \frac{F_{s.f.}}{mg} = \frac{29155}{6000 \cdot 10} \approx 0.49$$

Answer: 29155 N and 0.49 correspondingly.