Temperature and Heat

1. The drawing shows a hydraulic system used with disc brakes. The force F is applied perpendicularly to the brake pedal. The brake pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius = $9.50*10^{-3}m$) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii = $1.90*10^{-2}m$), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of F is 9.00 N. Assume that the input piston and the output plungers are at the same vertical level and find the force applied to each side of the revolving disc.

Show My Homework - TH1

2. The brass bar and the aluminum bar in the drawing are each attached to an immovable wall. At 28°C the air gap between the rods is $1.3 10^{-3} m$. At what temperature will the gap be closed?

Show My Homework - TH2

3. A thin spherical shell of silver has an inner radius of $2.0 x 10^{-2}m$ when the temperature is 18 degrees Celsius. The shell is heated to 147 degrees Celsius. Find the changes in the interior volume of the shell.

1. Let $F$ the force applied by the foot to the pedal and $F1$ the force applied by the pedal to the master cylinder. The momentum of the two forces with respect to the axis is equal.
$F *0.1 =F1*0.05$
$F1 =2*F$
let $F2$ the force applied to one side of the revolving disk, S1 the surface area of the master cylinder and S2 the surface area of one of the 2 lateral plungers (cylinders). The pressure applied to the master cylinder is equal to the total pressure of the plungers.
$P1 =2*P2$
$F1/S1 =2*F2/S2$
$F2 =F1 *S2/(2*S1) =2*F*R2^2/(2*R1^2) =2*9 *1.9^2*10^{-4}/(2*9.5^2*10^{-6}) = 36 N$

2. The linear dilatation of a rod obeys the law
$\Delta(L) = L*\alpha*\Delta(T)$
where $\alpha_{brass} =19*10^{-6} (Celsius)^{-1}$, $\alpha_{Aluminum} =23*10^{-6} (Celsius)^{-1}$
$\Delta(L_{brass}) +\Delta(L_{Al}) =1.3*10^{-3} m$
$L_{brass}*\alpha_{brass}*\Delta(T) +L_{Al}*\alpha_{Al}*\Delta(T) =1.3*10^{-3}$
$(2*19*10^{-6} +1*23*10^{-6})*\Delta(T) =1.3*10^-{3}$
$61*10^{-6}*delta(T) =1.3*10^{-3}$
$\Delta(T) = 21.311$ degree Celsius
$T = 28+delta(T) =28+21.3 =49.3$ degree Celsius

3. The variation of the surface S of the shell is obeying the law
$\delta(S) =S*\alpha_{Surface}*\Delta(T)$
$\alpha_{silverSurface} = 2*\alpha_{silverLiniar} =2*18*10^{-6} (Celsius)^{-1} =36*10^{-6} (Celsius)^{-1}$
Initial $S0 = 4*pi*R0^2 =4*pi*2*2*10^{-4} =50.26*10^{-4} m^2$
$\Delta(T) =147-18=129$ deg Celsius
$\Delta(S) =50.26*10^{-4}*36*10^{-6}*129 =2.33*10^{-5} m^2$
Final $S =S0 +\Delta(S) =(50.26+0.233)*10^{-4} =50.49*10^{-4} m^2$
The final radius of the shell is
$R =\sqrt{S/(4\pi)} =\sqrt{50.49*10^{-4}/(4\pi)} =2.0044*10^{-2} m$
The change in the interior volume is $V-V0 =(4*\pi)/3 *(R^3-R0^3) =(4*\pi)/3 *(2.0044^3*10^{-6} -2^3*10^{-6}) =2.2*10^{-7} m^3$