Theory Of Machine - Study Mode
[#156] Two heavy rotating masses are connected by shafts of lengths $${l_1},,{l_2}$$ xa0and $${l_3}$$ and the corresponding diameters are d 1 , d 2 and d 3 . This system is reduced to a torsionally equivalent system having uniform diameter d = d 1 of the shaft. The equivalent length of the shaft is
Correct Answer
(C) $${l_1} + {l_2}{left( {frac{{{{ ext{d}}_1}}}{{{{ ext{d}}_2}}}}
ight)^4} + {l_3}{left( {frac{{{{ ext{d}}_1}}}{{{{ ext{d}}_3}}}}
ight)^4}$$
[#157] The radius of a friction circle for a shaft rotating inside a bearing is (where r = Radius of shaft and $$ an varphi $$ xa0= Coefficient of friction between the shaft and bearing)
Correct Answer
(A) $${ ext{r}}sin varphi $$
[#158] Any point on a link connecting double slider crank chain will trace a
Correct Answer
(C) Ellipse
[#159] The method of obtaining different mechanisms by fixing in turn different links in a kinematic chain, is known as
Correct Answer
(C) Inversion
[#160] In considering friction of a V-thread, the virtual coefficient of friction (μ 1 ) is given by
Correct Answer
(D) $${mu _1} = frac{mu }{{cos x08eta }}$$
Explanation
Solution: The virtual coefficient of friction in a V-thread is derived from the normal reaction forces and frictional forces acting on the inclined plane of the thread. In a V-thread, the friction force is influenced by the helix angle and the thread angle ((x08eta)). The relationship between the actual coefficient of friction ((mu)) and the virtual coefficient of friction ((mu_1)) is determined by resolving forces along and perpendicular to the inclined thread surface. For a V-thread, the normal force component acts at an angle, leading to the effective coefficient of friction being altered by a factor of ( cos x08eta ). Thus, the virtual coefficient of friction is given by: [
mu_1 = frac{mu}{cos x08eta}
] This indicates that the effective friction increases as the thread angle ((x08eta)) increases because the normal force component decreases, requiring a larger effective friction to compensate.