Question

F sin 30:{}.30- 30° 300 Is Fcos 30° Nw = mg mig Free-body diagram Example 11: A crate is pulled by a force Fas shown. What is the minimum magnitude E of force needed to move the crate? Take m = 40 kg, us = 0.65, Mk = 0.5. Once the crate is moving under this force F what is the acceleration? Assume F is constant.

Answer 1

When the crate is just about to move, the frictional force acting on it is the static frictional force and the acceleration is zero

Net Force = mass*acceleration =0

Net Force = 0

$Fcos30-f_{s}=0$

$Fcos30-\mu _{s}N=0$

$Fcos30-\mu _{s}(mg-Fsin30)=0$

$Fcos30=\mu _{s}(mg-Fsin30)$

$Fcos30=\mu _{s}mg-\mu _{s}Fsin30$

$Fcos30+\mu _{s}Fsin30=\mu _{s}mg$

$F(cos30+\mu _{s}sin30)=\mu _{s}mg$

$F=\frac{\mu _{s}mg}{(cos30+\mu _{s}sin30)}$

$F=\frac{0.65*40kg*9.81m/s^{2}}{(cos30+0.65*sin30)}$

$F=\frac{0.65*40*9.81}{(cos30+0.65*sin30)}$

ANSWER: ${\color{Red} F=214.1516N}$

============================

When the crate is moving, the frictional force acting on it is the kinetic frictional force.

Net Force = mass*acceleration

$Fcos30-f_{k}=ma$

$Fcos30-\mu _{k}N=ma$

$Fcos30-\mu _{k}(mg-Fsin30)=ma$

$214.1516N*cos30-0.5*(40kg*9.81m/s^{2}-214.1516N*sin30)=40kg*a$

$214.1516*cos30-0.5*(40*9.81-214.1516*sin30)=40*a$

$214.1516*cos30-0.5*285.3242=40*a$

$\frac{214.1516*cos30-0.5*285.3242}{40}=a$

ANSWER: ${\color{Red} a=1.07m/s^{2}}$

======================