For the 200-kg space vehicle of Sample Prob. 14.1, it is known that at t = 2.5 s, the velocity of part A is vA = (270 m/s)i - (120 m/s)j + (160 m/s)k and the velocity of part B is parallel to the xz plane. Determine the velocity of part C.

Question

For the 200-kg space vehicle of Sample Prob. 14.1, it is known that at t = 2.5 s, the velocity of part A is [latex] ext v_A[/latex] = (270 m/s)i - (120 m/s)j + (160 m/s)k and the velocity of part B is parallel to the xz plane. Determine the velocity of part C.

Accepted Answer

Since there is no external force, the initial momentum m[latex] ext v_0[/latex] is equipollent to the system of the final momenta. Equating first the sums of the vectors in both parts of the adjoining sketch, and then the sums of their moments about O, we write

[latex]L _1= L _2: \quad\quad m ext v _0=m_A ext v _A+m_B ext v _B+m_C ext v _C[/latex] (1)

[latex]\left( H _O ight)_1=\left( H _O ight)_2: \quad\quad 0= r _A imes m_A ext v _A+ r _B imes m_B ext v _B+ r _C imes m_C ext v _C[/latex] (2)

Recalling from Sample Prob. 14.1 that [latex] ext v _0=(150 m / s ) i[/latex]

[latex]m_A=100 ~kg \quad m_B=60~ kg \quad m_C=40~ kg[/latex]

[latex]r _A=(555~ m ) i -(180~ m ) j +(240~ m ) k[/latex]

[latex]r _B=(255 ~m ) i -(120 ~m ) k[/latex]

[latex]r _C=(105~ m ) i +(450 ~m ) j -(420 ~m ) k[/latex]

and using the information given in the statement of this problem, we rewrite Eqs. (1) and (2) as follows:

[latex]200(150 i )=100(270 i -120 j +160 k )+60\left[\left(v_B ight)_x i +\left(v_B ight)_z k ight] \ \ +40\left[\left(v_C ight)_x i +\left(v_C ight)_y j +\left(v_C ight)_z k ight][/latex] (1')

[latex]0=100\left|\begin{array}{ccc} i & j & k \555 & -180 & 240 \270 & -120 & 160\end{array} ight|+60\left|\begin{array}{ccc} i & j & k \255 & 0 & -120 \\left(v_B ight)_x & 0 & \left(v_B ight)_z\end{array} ight| \ \ +40\left|\begin{array}{ccc} i & j & k \105 & 450 & -420 \\left(v_C ight)_x & \left(v_C ight)_y & \left(v_C ight)_z\end{array} ight|[/latex] (2')

Equating to zero the coefficient of j in (1') and the coefficients of i and k in (2'), we write, after reductions, the three scalar equations

[latex]\left(v_C ight)_y-300=0[/latex]

[latex]450\left(v_C ight)_z+420\left(v_C ight)_y=0[/latex]

[latex]105\left(v_C ight)_y-450\left(v_C ight)_x-45~000=0[/latex]

which yield, respectively,

[latex]\left(v_C ight)_y=300 \quad\left(v_C ight)_z=-280 \quad\left(v_C ight)_x=-30[/latex]

The velocity of part C is thus

[latex]v _C=-(30 ~m / s ) i +(300~ m / s ) j -(280~ m / s ) k[/latex]

Question 14.3: For the 200-kg space vehicle of Sample Prob. 14.1, it is kno......

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