How much has the cable stretched?

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The discussion focuses on calculating the stretch of a steel cable being used to pull a car from a ditch. The original length of the cable is 9.1 meters, with a radius of 0.5 cm, and a tension of 890 N is applied. The formula used for calculating the change in length is correct, incorporating the Young's modulus for steel. The calculated stretch of the cable is approximately 5.2E-4 meters. The conversation concludes with a light-hearted comment about the type of pie, but the main topic remains the cable's stretch calculation.
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A truck is pulling a car out of a ditchwith a steel cable that is 9.1 meters long and has a radius of .5cm. When the car just begins to move the tension of the cable is 890N. How much has the cable stretched?
converted .5cm to .005m
area of the wire A = pie*r^2
Y = 20E10 (steel)
used the formula Change in L = F*orginal length/YA

Change in L = 890N*9.10m / 20E10* (pie*.005^2)

and got 5.2E-4 m

did i go about this correctly?

thanks joe
 
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Looks OK to me.
 
Originally posted by cowgiljl
A truck is pulling a car out of a ditchwith a steel cable that is 9.1 meters long and has a radius of .5cm. When the car just begins to move the tension of the cable is 890N. How much has the cable stretched?
converted .5cm to .005m
area of the wire A = pie*r^2
Y = 20E10 (steel)
used the formula Change in L = F*orginal length/YA

Change in L = 890N*9.10m / 20E10* (pie*.005^2)

and got 5.2E-4 m

did i go about this correctly?

thanks joe

Now, is this cable apple, or cherry pie?

I'm sorry I couldn't resist
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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