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alex41 [277]
3 years ago
15

A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer

?
O The nail exerts a much smaller force on the hammer in the opposite direction
O The nail exerts a much smaller force on the hammer in the same direction.
The nail exerts an equal force on the hammer in the same direction.
O The nail exerts an equal force on the hammer in the opposite direction.
Physics
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:

reviewing the final statements, the correct one is the quarter

The nail exerts an equal force on the hammer in the opposite direction.

Explanation:

This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.

This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.

The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of ​​the hammer is much greater, therefore the pressure is small.

When reviewing the final statements, the correct one is the quarter

The nail exerts an equal force on the hammer in the opposite direction.

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                        N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )

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                        F = u*N\\\\F = u*M*g *cos ( Q )

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

                       M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q)  = M*a\\ .. Eq2

- Add the two equation, Eq 1 and Eq 2:

                      M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}

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- Using equation 1, we determine the tension ( T ) in the cable as follows:

                     T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N

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