Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑
= mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
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Explanation:
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Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence
![T_{BE}=W=20*9.81=196.2 N](https://tex.z-dn.net/?f=T_%7BBE%7D%3DW%3D20%2A9.81%3D196.2%20N)
Therefore, tension in the cable, ![T_{BE}=196.2 N](https://tex.z-dn.net/?f=T_%7BBE%7D%3D196.2%20N)
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then
![196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0](https://tex.z-dn.net/?f=196.2%5Ctimes%200.125-%20196.2%5Ctimes%200.2%2B%20D_x%5Ctimes%200.2%3D0)
![24.525-39.24+0.2D_x=0](https://tex.z-dn.net/?f=24.525-39.24%2B0.2D_x%3D0)
![D_x=73.575 N](https://tex.z-dn.net/?f=D_x%3D73.575%20N)
Similarly,
![A_x-D_y=0](https://tex.z-dn.net/?f=A_x-D_y%3D0)
![A_x=73.575 N](https://tex.z-dn.net/?f=A_x%3D73.575%20N)
Therefore, both reactions at A and D are 73.575 N
Answer:
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Shipwrights — Shipwrights build, design, and repair all sizes of boats.
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