Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
the force of gravity between them is quadrupled .
Explanation:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.
To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.
In this way,
The total mass of the cars would be,


Therefore the friction force at 29Km / h would be,




In this way the tension exerts between first car and locomotive is,



Therefore the tension in the coupling between the car and the locomotive is 
Answer: 1.2m/s^2
Explanation: the force exerted on the car is 900N upwards
The mass of the car is 750kg
According to Newton's third law acceleration is proportional to force
F = ma
900 = 750a
a = 900/750
a = 1.2m/s^2
Answer:
9.51
Explanation:
The distance s is given by:

The change in distance is given by the time derivative of s:

For the time t you solve the equation of distance x for time:

Plugging in for t:
