A worker wants to turn over a uniform, 1250-N, rectangular crate by pulling aton one of its vertical sides. The floor is rough enough to prevent the crate from slipping.

1. What pull is needed to just start the crate to tip?
2. How hard does the floor push upward on the crate?
3. Find the friction force on the crate.
4. What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

Given that a worker wants to turn over a uniform, 1250-N, rectangular crate by pulling at$53.0°$ on one of its vertical sides. The floor is rough enough to prevent the crate from slipping.

Weight of rectangular crate w = 1250N

Length of crate I = 2.2 m

Breadth of crate, b = 1.5 m

The bar is in equilibrium until the cable breaks, so the forces and torques must balance. Hence condition for equilibrium can be applied.

Consider the formula for the torque is

$\mathbf{\tau }\mathbf{=}\mathbf{FI}$

Here, Fis force exerted and l is moment arm.

Let the pull needed to start the crate be P

According to condition of equilibrium about the center of the crate

Net torque is zero.

So $P\left(b\right)\sin 53°-w\left(\frac{I}{2}\right)=0$

$\begin{array}{r}P\left(1.5\mathrm{m}\right)\sin {53}^{\circ }-\left(1250\mathrm{N}\right)\left(1.1\mathrm{m}\right)=0\\ P=\frac{\left(1250\mathrm{N}\right)\left(1.1\mathrm{m}\right)}{\left(1.5\mathrm{m}\right)\sin {53}^{\circ }}=1150\mathrm{N}\end{array}$

Hence, force needed to start the crate$1.15\times {10}^3\mathrm{N}$

Let H be the force acting on crate by floor.

Let$H_x$be horizontal force and$H_y$be vertical force.

Now, according to condition of equilibrium,

Net force is zero.

So this gives$\sum _{}{F}_x=0$and$\sum _{}{F}_y=0$

Now, the floor pushes upwards implies vertical force, so

$\begin{array}{r}\sum F_y=0\text{implies}{H}_y-w-P\cos {53}^{\circ }=0\\ H_y=\left(1250\mathrm{N}\right)+\left(1150\mathrm{N}\right)\cos {53}^{\circ }=1940\mathrm{N}\end{array}$.

Hence, floor push crate upward by a force of $1.94\times {10}^3\mathrm{N}$

Let H be the force acting on crate by floor.

Let $H_x$be horizontal force and$H_y$ be vertical force.

Now, according to condition of equilibrium,

Net force is zero.

So this gives$\sum _{}{F}_x=0and\sum _{}{F}_y=0$ and .

Now, frictional force acts horizontally. so

$\sum _{}{F}_x=0$implies $H_x-P\sin 53°=0$

$H_x=0implies{H}_x-P\sin 53°=0$

Hence, frictional force on the crate is 918 N

The coefficient of static friction is $\mu$.

Now, $\mu =\frac{H_x}{H_y}$

So $\mu =\frac{918\mathrm{N}}{1940\mathrm{N}}=0.473$

Hence, coefficient of static friction is 0.473.