If tension in the cable supporting a lift moving downwards is half the tension when it is moving upwards, the acceleration of the lift is
A. $$\frac{{\text{g}}}{2}$$
B. $$\frac{{\text{g}}}{3}$$
C. $$\frac{{\text{g}}}{4}$$
D. None of these
Answer: Option D
This Question Belongs to Mechanical Engineering >> Engineering Mechanics
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The resultant of two equal forces P making an angle $$\theta ,$$ is given by
A. $$2{\text{P}}\sin \frac{\theta }{2}$$
B. $$2{\text{P}}\cos \frac{\theta }{2}$$
C. $$2{\text{P}}\tan \frac{\theta }{2}$$
D. $$2{\text{P}}\cot \frac{\theta }{2}$$
A. Equal to
B. Less than
C. Greater than
D. None of these
If a number of forces are acting at a point, their resultant is given by
A. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2}$$
B. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2}} $$
C. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)$$
D. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)} $$
A. $${\text{a}} = \frac{\alpha }{{\text{r}}}$$
B. $${\text{a}} = \alpha {\text{r}}$$
C. $${\text{a}} = \frac{{\text{r}}}{\alpha }$$
D. None of these
Mg-Ma=1/2(Ma-Mg),
2Mg-2Ma=Ma-Mg,
3Mg=3Ma,
a=g
b g/3 hoga
B... g/3 hoga
Formula for Tension (T) = mg +ma
where:
T=Tension
m= mass of an object
a= acceleration
therefore:
1/2 = mg + ma
ma= 2mg, where m= constant (cancelled)
Acceleration (a) = 2g
ANSWER: D (none of the above)
T = m x g + m x a
where:
T = Tension
m = mass of body
g = gravitational force
a = acceleration
therefore:
T = 1/2
1/2 = mg + ma
ma = 2mg
a =2g
Answer: D (None of These)
It's B.
1/2 upward = Downward.
1/2 m(g+a) = mg-ma,
mg+ma = 2mg-2ma,
mg =3ma,
g = 3a,
a = g/3.
B rat ko
A din me
C do peher ko
D sapne me
Lift moving up
T=mg +ma
Lift moving down
T=mg-ma
T down = 0.5 T up
0.5 mg+0.5ma=mg-ma
a=g/3