Answer to Question #260913 in Mechanics | Relativity for Celina Keays

Question

A box has a weight of 5000N. If the coefficient of static friction is 0.8 and the coefficient of kinetic friction is 0.5. How much force is required to get the box to start moving? How much is required to move the box at a constant velocity?

Answer 1

$\vec P = 5000N$

$\mu_{sf}=0.8$

$\mu _{kf}= 0.5$

$\vec{N}=-\vec P$

$\text{for static friction:}$

$F{sf}=\mu_{sf} |\vec N|= 0.8*5000=4000N$

$F_1>4000N -\text{the force required for the box to start moving}$

$\text{for kinetic friction:}$

$\vec F_2 -\vec F_{kf}= m\vec a$

$\vec v = const ,\text{ then }\vec a=0$

$\vec F_2 =\vec F_{kf}$

$\vec F_{kf}=\mu_{kf}|\vec N|= 0.5*5000N=2500N$

$\vec F_2=2500N,\text{force required to move the box evenly}$

$\text{Answer:}$

$F_1>4000N ,\text{the force required for the box to start moving}$

$\vec F_2=2500N,\text{force required to move the box evenly}$