Q11-35E - University Physics with Modern Physics

Question

A copper cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. (Consult Table 11.1.) (a) Show that the glue exerts a force F on the bottom face that is equal in magnitude but opposite to the force on the top face. (b) How large must F be to cause the cube to deform by 0.250 mm? (c) If the same experiment were performed on a lead cube of the same size as the copper one, by what distance would it deform for the same force as in part (b)?

Step-by-Step Solution

Answer

(a) Glue exerts a force F in the opposite direction

(b) $6.6\times {10}^{5}N$

(c) $1.8 mm$

Step 1: Given information:

$Cube side a=6cm, Area A={\left(6cm\right)}^2, from Table 11.1 S=4.4\times {10}^{10}Pa$

Step 2: Concept/Formula used:

The deformation x is related to the force by

$\mathit{S}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{I}\mathbf{I}}\mathbf{a}}{\mathbf{A}\mathbf{x}}$

Where, 

F_II is force parallel to surface area

Step 3: Force exerted by glue

(a) The forces on the cube must balance. Since the horizontal forces balance, the glue exerts a force F in the opposite direction. 

 

(b)

$\begin{array}{rcl}F& =& \frac{AxS}{h}\\ & =& \frac{{\left(0.06m\right)}^2\left(0.250\times {10}^{-3}m\right)\left(4.4\times {10}^{10}Pa\right)}{0.06m}\\ & =& 6.6\times {10}^{5}N\end{array}$

Step 4: Deformation on lead cube

(c)

$\begin{array}{rcl}x& =& \frac{F_{II}h}{AS}\\ & =& \frac{\left(6.6\times {10}^{5}N\right)\left(0.06m\right)}{{\left(0.06m\right)}^2\left(0.6\times {10}^{10}Pa\right)}\\ & =& 1.8mm\end{array}$