Question 1 |
The braking system shown in the figure uses a belt to slow down a pulley rotating
in the clockwise direction by the application of a force P. The belt wraps around
the pulley over an angle \alpha = 270 degrees. The coefficient of friction between the
belt and the pulley is 0.3. The influence of centrifugal forces on the belt is
negligible.
During braking, the ratio of the tensions T_1 to T_2 in the belt is equal to __________. (Rounded off to two decimal places)
Take \pi =3.14.
During braking, the ratio of the tensions T_1 to T_2 in the belt is equal to __________. (Rounded off to two decimal places)
Take \pi =3.14.
2.12 | |
6.25 | |
4.11 | |
8.25 |
Question 1 Explanation:
\begin{aligned}
\mu & =0.3 \\
\alpha & =270^{\circ}=\frac{3 \pi}{2} \\
\frac{T_{1}}{T_{2}} & =\mathrm{e}^{\mu \alpha}
\end{aligned}
So, \quad \frac{T_{1}}{T_{2}}=e^{0.3 \times \frac{3 \pi}{2}}=4.11
So, \quad \frac{T_{1}}{T_{2}}=e^{0.3 \times \frac{3 \pi}{2}}=4.11
Question 2 |
A short shoe drum (radius 260 mm) brake is shown in the figure. A force of 1 kN is applied to the lever. The coefficient of friction is 0.4.
The magnitude of the torque applied by the brake is ________N.m (round off to one decimal place).
The magnitude of the torque applied by the brake is ________N.m (round off to one decimal place).
150 | |
175 | |
200 | |
250 |
Question 2 Explanation:
Taking moment about 'O'
\begin{aligned} R_{N}(500)+F_{\gamma}[310-260] &-1000 \times 1000=0 \\ R_{N}(500)+0.4\left(R_{N}\right)(50) &-1000 \times 1000=0 \\ R_{N} &=1923.076 \mathrm{~N} \\ F_{r} &=\mu R_{N}=769.23 \mathrm{~N} \\ T_{f} &=F_{r} \times R=200 \mathrm{~N}-\mathrm{m} \end{aligned}
Question 3 |
A helical spring has spring constant k. If the wire diameter, spring diameter and the
number of coils are all doubled then the spring constant of the new spring becomes
k/2 | |
k | |
8k | |
16k |
Question 3 Explanation:
\begin{aligned} k_{(\text {sping })} &=\frac{G d^{4}}{8 D^{3} n} \\ k_{\text {new }} &=\frac{G(2 d)^{4}}{8(2 D)^{3}(2 n)}=\frac{G d^{4}}{8 D^{3} n} \\ \text{Hence}\qquad k_{\text {new }} &=k \end{aligned}
Question 4 |
In a disc-type axial clutch, the frictional contact takes place within an annular region
with outer and inner diameters 250 mm and 50 mm, respectively. An axial force F_1 is
needed to transmit a torque by a new clutch. However, to transmit the same torque,
one needs an axial force F_2 when the clutch wears out. If contact pressure remains uniform
during operation of a new clutch while the wear is assumed to be uniform for an old
clutch and the coefficient of friction does not change, then the ratio F_1/F_2 is_________
(round off to 2 decimal places).
0.22 | |
0.46 | |
0.87 | |
0.93 |
Question 4 Explanation:
\begin{aligned} T &=\mu W \times R_{m} \\ T_{\text {new }} &=\mu \times W \times R_{\text {new }} \\ T_{\text {new }} &=\mu \times W_{\text {new }} \times \frac{2}{3} \frac{\left(R_{0}^{3}-R_{i}^{3}\right)}{R_{0}^{2}-R_{i}^{2}} \\ T_{\text {odd }} &=\mu \times W \times R_{\text {old }} \\ T_{\text {new }} &=T_{\text {dd }} \\ W_{\text {new }} \times \frac{2}{3} \frac{\left(R_{0}^{3}-R_{i}^{3}\right)}{R_{0}^{2}-R_{i}^{2}} &=W_{\text {old }} \times \frac{\left(R_{0}+R_{i}\right)}{2} \\ \frac{W_{\text {new }}}{W_{\text {old }}} &=0.871 \end{aligned}
Question 5 |
A single block brake with a short shoe and torque capacity of 250 N\cdot m is shown. The cylindrical brake drum rotates anticlockwise at 100 rpm and the coefficient of friction is 0.25. The value of a, in mm (round off to one decimal place), such that the maximum actuating force P is 2000 N, is ________
212.5 | |
256.4 | |
159.8 | |
753.5 |
Question 5 Explanation:
\mu=0.25
76 Torque =250 \mathrm{Nm}
\mu \mathrm{F} \times \mathrm{a}=250
\Rightarrow 0.25 \times \mathrm{F} \times \mathrm{a}=250
\Rightarrow \mathrm{F} \times \mathrm{a}=1000
In FBD of block and lever, \Sigma \mathrm{M}_{\text {fullcrum }}=0
\mathrm{F} \times \mathrm{a}+\mu \mathrm{F} \times \frac{\mathrm{a}}{4}=\mathrm{P} \times 2.5 \mathrm{a}
F \times a\left(1+\frac{\mu}{4}\right)=P \times 2.5
1000\left(1+\frac{0.25}{4}\right)=2000 \times 2.5 \mathrm{a}
\Rightarrow \mathrm{a}=0.2125 \mathrm{m}=212.5 \mathrm{mm}
There are 5 questions to complete.