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A baseball of mass m = 0.16 kg is suspended vertically from a tree by a string of length L = 1.4 m and negligible mass. Take z as the upward vertical direction. A short-duration force F = Fyj + Fzk is applied to the baseball by a bat, momentarily creating a torque τ about the top of the string.Enter an expression for the torque due to the given force about the top of the string, in terms of m, L, Fy F , and the unit vectors i

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Answer:

[tex]\tau = L F_y \hat i [/tex]

Explanation:

As we know that torque due to applied force is given as

[tex]\tau = \vec r \times \vec F[/tex]

here we know that

[tex]\vec r = -L\hat k[/tex]

applied force is given as

[tex]\vec F = F_y \hat j + F_z \hat k[/tex]

now we will have

[tex]\tau = (-L \hat k) \times (F_y \hat j + F_z \hat k)[/tex]

[tex]\tau = L F_y \hat i [/tex]

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