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प्रश्न
A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in figure. Match the following (most appropriate choice):
| (a) mg/4 < F < mg/2 | (i) Cube will move up. |
| (b) F > mg/2 | (ii) Cube will not exhibit motion. |
| (c) F > mg | (iii) Cube will begin to rotate and slip at A. |
| (d) F = mg/4 | (iv) Normal reaction effectively at a/3 from A, no motion. |
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उत्तर
| (a) mg/4 < F < mg/2 | (ii) Cube will not exhibit motion. |
| (b) F > mg/2 | (iii) Cube will begin to rotate and slip at A. |
| (c) F > mg | (i) Cube will move up. |
| (d) F = mg/4 | (iv) Normal reaction effectively at a/3 from A, no motion. |
Explanation:
Consider the below diagram
Moment of the force F about point A, τ1 = F × a .....(anti-clockwise)
Moment of weight mg of the cube about point A.
τ2 = `mg xx a/2` .....(clockwise)
Cube will not exhibit motion, If τ1 = τ2 ......(∵ In this case, both the torque will cancel the effect of each other)
∴ F × a = `mg xx a/2`
⇒ F = `(mg)/2`
Cube will rotate only when, τ1 > τ2
⇒ F × a > `mg xx a/2`
⇒ `F > (mg)/2`
Let the normal reaction is acting at `a/3` from point A, then
`mg xx a/3 = F xx a` or `F = (mg)/3` .......(For no motion)
When F = `(mg)/4` which is less than `(mg)/3`, .....`(F < (mg)/3)`
There will be no motion.
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