Question

Question: A uniform cubical crate is 0.750 m on each side and weighs 500 N . It rests on a floor with one edge against a very small, fixed obstruction. At what least height above the floor must a horizontal force of magnitude 350 N be applied to the crate to tip it?

Short answer

Answer:

A horizontal force of 350 N must be applied at to tip the crate.

Step-by-step solution

Understanding the given information   

$$\begin{gathered} a=0.75\text{m} \\ W=500\text{N} \\ F=350\text{N} \end{gathered}$$

Concept and formula used in the given question

Using the conditions for static equilibrium, you can write the equation for torque in terms of force and distance. Using this equation, you can solve for unknown height h. The equations used are given below.

Static equilibrium conditions:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

Calculation for the at least what height above the floor must a horizontal force of magnitude  350 N be applied to the crate to tip it  

Applying equilibrium condition:

$$\begin{gathered} \sum \tau =0 \\ W\times \left(\frac{a}{2}\right)-F\times h=0 \\ 500\times \left(\frac{0.75}{2}\right)-\left(350\times h\right)=0 \\ h=0.536\text{m} \end{gathered}$$

Hence,a horizontal force of 350 N must be applied at$h=0.536\text{m}$ to tip the crate.