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Tutor profile: Christodoulos K.

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Christodoulos K.
Friendly & competent tutor with expertise in Applied Maths & Physics
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

A mass, $$m=14kg$$, is pulled with a force P, on an inclined slope of $$w=25degrees$$ with an acceleration of $$a=2.6m/s^2$$. The coefficient of friction is$$ f=0.18$$. Find force P.

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Christodoulos K.
Answer:

Draw out a Free Body Diagram to acquire the following equilibrium equations (where N = normal force, F = Friction force): $$F+ma+mgsin(w)=P$$ $$N = mgcos(w)$$ Thus: $$N = 124.47N$$ Using the Coulomb model of friction: $$F = fN=0.18\times124.47N = 22.41N$$ Hence, using the first equilibrium equation, P is found to be: $$P=116.85N$$.

Subject: Mechanical Engineering

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Question:

What is the creep phenomenon exhibited by materials and is this phenomenon common in metals?

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Christodoulos K.
Answer:

In Materials Science, the time dependent deformation of a material that is subjected to a constant load is known as creep. Unlike perfectly elastic (or Hookean) behavior that describes an instant deformation upon application of a load, a material that undergoes creep can gradually deform over a long period of time. Creep is common in Viscoelastic materials (most polymers) and is also common in metals at elevated temperatures (e.g. "super-alloy" turbine blades found in jet-engines).

Subject: Differential Equations

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Question:

A 1kg mass, m, with initial displacement, x = 10mm, is connected to a shock absorber with spring stiffness k=24N/mm and damping coefficient c=0.2Ns/mm. Using the Second Ordinary Differential Equation of a free mechanical system as $$ \frac{d^2x}{dt^2} + \frac{c}{m} \frac{dx}{dt}+ \frac{k}{m} x=0$$ find the particular solution of the system.

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Christodoulos K.
Answer:

By substituting values for k,c and m, we get the following expression: $$ \frac{d^2x}{dt^2} + 0.2 \frac{dx}{dt}+ 24 x=0$$, given that x(0) = 0.1m and $$ \frac{dx}{dt}=0 $$ at t=0. The charachteristic equation is: $$ m^2 +\frac{m}{5}+ 24 =0$$ so that m = 0.1+4.90j and m = -0.1-4.90j. The general solution can thus be found as: $$ x(t) = exp(-0.1t) (Acos(4.9t)+Bsin(4.9t))$$ Applying the two initial conditions x(0) = 0.1m and $$ \frac{dx}{dt}=0 $$ we arrive at the particular solution: $$ x(t) = exp(-0.1t) (10cos(4.9t)+0.2sin(4.9t))$$.

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