In either of these situations, we define stress as the ratio of the deforming force
${F}_{\perp}$ to the cross-sectional area
A of the object being deformed. The symbol
${F}_{\perp}$ that we reserve for the deforming force means that this force acts perpendicularly to the cross-section of the object. Forces that act parallel to the cross-section do not change the length of an object. The definition of the tensile stress is
$\text{tensile stress}=\frac{{F}_{\perp}}{A}.$
Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional change of the object’s length when the object experiences tensile stress
Compressive stress and strain are defined by the same formulas,
[link] and
[link] , respectively. The only difference from the tensile situation is that for compressive stress and strain, we take absolute values of the right-hand sides in
[link] and
[link] .
Young’s modulus
Y is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by
[link] . Dividing this equation by tensile strain, we obtain the expression for Young’s modulus:
A sculpture weighing 10,000 N rests on a horizontal surface at the top of a 6.0-m-tall vertical pillar
[link] . The pillar’s cross-sectional area is
$0{\text{.20 m}}^{2}$ and it is made of granite with a mass density of
${2700\phantom{\rule{0.2em}{0ex}}\text{kg/m}}^{3}.$ Find the compressive stress at the cross-section located 3.0 m below the top of the pillar and the value of the compressive strain of the top 3.0-m segment of the pillar.
Strategy
First we find the weight of the 3.0-m-long top section of the pillar. The normal force that acts on the cross-section located 3.0 m down from the top is the sum of the pillar’s weight and the sculpture’s weight. Once we have the normal force, we use
[link] to find the stress. To find the compressive strain, we find the value of Young’s modulus for granite in
[link] and invert
[link] .
Solution
The volume of the pillar segment with height
$h=3.0\phantom{\rule{0.2em}{0ex}}\text{m}$ and cross-sectional area
$A=0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ is
With the density of granite
$\rho =2.7\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3},$ the mass of the pillar segment is
The weight of the sculpture is
${w}_{s}=1.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N},$ so the normal force on the cross-sectional surface located 3.0 m below the sculpture is
Young’s modulus for granite is
$Y=4.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\text{Pa}=4.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\text{kPa}.$ Therefore, the compressive strain at this position is
Notice that the normal force acting on the cross-sectional area of the pillar is not constant along its length, but varies from its smallest value at the top to its largest value at the bottom of the pillar. Thus, if the pillar has a uniform cross-sectional area along its length, the stress is largest at its base.
net force is when you add forces numerically I.e. the total sum of all positive and negative or balanced and unbalanced forces.
resultant force is a single vector which is the combination or addition of all x and y axes vector component forces in a system.
emmanuel
thanks
Ogali
resultant force is applied to hold or put together an object moving at the wrong direction. in other words it repairs.
Andrew
Damping is provided by tuning the turbulence levels in the moving water using baffles.How it happens? Give me a labelled diagram of it.
A 10kg ball travelling at 4meter per second collides elastically in a head-on collision with a 2kg ball.What are (a)the velocities and (b)the total momentum of the balls after collision?
a)v1 8/3s&v2 20/3s. b)in elastic collision total momentum is conserved.
Bala
multiply both weight which is 10*2 divided by the time give 4. and our answer will be 5.
Andrew
The displacement of the air molecules in sound wave is modeled with the wave function s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t)s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t) . (a) What is the wave speed of the sound wave? (b) What is the maximum speed of the air molecules as they oscillate in simple harmon
If the block is displaced to a position y , the net force becomes Fnet=k(y−y0)−mg=0Fnet=k(y−y0)−mg=0 . But we found that at the equilibrium position, mg=kΔy=ky0−ky1mg=kΔy=ky0−ky1 . Substituting for the weight in the equation yields. Show me an equation of graph.