calculate the magnitude of contact force applied by 4kg mass on 6 kg mass shown in figure
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gangester10eeee:
hlo mere saat Chat karogi meri gf banogi meri gf banogi meri gf
apka naam
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Answered by
3
According to the free body diagrams of each of these masses
The following equations can be written :
1) 30 - N = 4a
and 2) N = 6a
So,
a = N/6
Thus,
30 - N = 4*(N/6)
-> 30 = 5N/3
Therefore,
N = 18 Newton
Therefore, the contact or normal interaction force between the blocks will be 18 Newton
Remember, the acceleration of both the blocks will be the same.
So, that they do not lose contact and the normal reaction force doesn't become equal to zero.
The following equations can be written :
1) 30 - N = 4a
and 2) N = 6a
So,
a = N/6
Thus,
30 - N = 4*(N/6)
-> 30 = 5N/3
Therefore,
N = 18 Newton
Therefore, the contact or normal interaction force between the blocks will be 18 Newton
Remember, the acceleration of both the blocks will be the same.
So, that they do not lose contact and the normal reaction force doesn't become equal to zero.
thanks
No problem
i have one more doubt..
yes?
a man pushes a body of mass 10kg with a force of 10N. the body pushes the man back with the same magnitude of force. is this statement is true or false..?
It is absolutely true, due to the Newton's Third Law, that interactive forces between two objects occur as an action reaction pair in universe.
ok thanks
Yep ｡◕‿◕｡
Answered by
2
the required answer is 18 Newton
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