During fundraising week at school, students decide to hold a competition of strength. Contestants...

Question:

During fundraising week at school, students decide to hold a competition of strength. Contestants pay $1 to try to move a heavy object. If they can move the object, they win $10. The object is a large box of books of mass 250 kg, with a coefficient of static friction 0.55. One student has a mass of 64 kg and a coefficient of static friction of 0.72 for his shoes on the floor.

a. What is the maximum force of static friction acting on the student?

b. What is the maximum force of static friction acting on the box?

c. Is the competition fair? Explain your reasoning.

Answer and Explanation:

Formula for static friction

The formula for solving the amount of static friction between an object of mass {eq}m {/eq} at rest and a surface where the object sits on is given by:

{eq}\begin{align*} f_S &= \mu_S N \end{align*} {/eq}

where {eq}\mu_S {/eq} is the coefficient of static friction and N is the amount of normal force. Since the object is on the surface with its own weight, W = mg, where {eq}g = 9.8 \mathrm{m/s^2} {/eq} is the gravitational acceleration, the normal force is given by:

{eq}\begin{align*} N &= W = mg \end{align*} {/eq}

Thus, by substituting this into the previous equation the formula for static friction can be rewritten as:

{eq}\begin{align*} f_S &= \mu_S mg \end{align*} {/eq}

(a) Static friction on the student

The static friction on the student of mass {eq}m_s = 64 \mathrm{kg} {/eq} by his shoes with coefficient of static friction {eq}\mu_{S,s}= 0.72 {/eq} using the formula is thus given by:

{eq}\begin{align*} f_{S,s} &= \mu_{S,s} m_s g\\ &= (0.72)(64 \ \mathrm{kg})(9.8 \ \mathrm{m/s^2})\\ &= 451.58 \ \mathrm{kg \ m/s^2} = 451.58 \ \mathrm{N} \end{align*} {/eq}

(b) Static friction on the box

For the books box, the static friction with its mass {eq}m_b = 250 \mathrm{kg} {/eq} and coefficient of static friction {eq}\mu_{S,b}= 0.55 {/eq} using the formula is thus given by:

{eq}\begin{align*} f_{S,b} &= \mu_{S,b} m_b g\\ &= (0.55)(250 \ \mathrm{kg})(9.8 \ \mathrm{m/s^2})\\ &= 1347.5 \ \mathrm{kg \ m/s^2} = 1347.5 \ \mathrm{N}\\ \end{align*} {/eq}

(c) Is the competition fair?

The static friction of the box and the student is the maximum force on each mass's contact on the floor that an external force should overcome in order to move that mass out of rest. In this competition, the student gets to push the box of books to win, so whatever force the student applies on the box would also be the same force experienced by the contact of the bottom of the box and the student's shoes. The student should exert more and more force until he overcomes the static friction of the box in order to get it moving. But in the case of this 64 kg student, the static friction of the box is greater than the static friction on his shoes, so that with the increasing force he exerts his shoes slide off the floor even before he can get to move the box. Under such circumstances this student will never be able to move the box at all.

The only thing this student could do is to wish that he became heavier or that he had a better pair of shoes with higher static friction coefficient so that he might be able to overcome the static friction of the box. Thus, the competition would never be fair for students of different mass or students with other pairs of shoes, even if all the students are assumed to be equally strong.


Learn more about this topic:

Static Friction: Definition, Formula & Examples

from High School Physical Science: Tutoring Solution

Chapter 11 / Lesson 21
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