Question

A small box of mass m₁ is sitting on a board of mass m₂ and length L. The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is μ_s. The coefficient of kinetic friction between the board and the box is, as usual, less than μ_s.

Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use F_f for the magnitude of the friction force between the board and the box.

Find F_min, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the box (which will then fall off of the opposite end of the board).

Express your answer in terms of some or all of the variables μ_s, m₁, m₂, g , and L. Do not include F_f in your answer.

Answer 1

Concepts and reason

The concepts required to solve this problem are Newton’s second law and friction force.

Initially, use the expression of static friction and Newton’s second law to solve for the minimum acceleration required to pull the board out.

Finally, use the Newton’s second law for the system of box and the board to solve for the minimum constant force required.

Fundamentals

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, $\sum F$ is the sum of all the forces on the object, $m$ is mass of the object, and $a$ is the acceleration of the object.

The expression for static friction over a horizontal surface is,

${F_{\rm{f}}} \le {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, $m$ is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass${m_1}$.

Substitute ${m_1}$ for $m$ in the expression of maximum static friction${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$.

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration $a$ to pull the box out.

Substitute ${F_{\rm{f}}}$ for$\sum F$, ${m_1}$ for$m$in the equation$\sum {F = ma}$.

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation${F_{\rm{f}}} = {m_1}a$.

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

Use the Newton’s second law for the system of box and the board.

Substitute ${F_{{\rm{min}}}}$ for$\sum F$, $\left( {{m_1} + {m_2}} \right)$ for$m$in the equation$\sum {F = ma}$.

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute ${\mu _{\rm{s}}}g$ for $a$ in the above equation${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$.

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

Ans:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is$\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$.