The intention is to determine whether the cables will resist the tension or will break.
There are three tensions
Applyng Newton's Second Law to the student, the tension of the only cable that holds the student has to equal his weight,
T = weight = m*g = 160 lbs / 2.2046 lbs/kg * 9.8 m/s=711 N
Now apply Newton's Second Law to the joint of the cables
There you have that the equilibrium of forces leads to that the sum of the up-components of the other two cables = the tension T just found, i.e. 711 N.
Now find the up-components of the tensions of other two cables:
sin 39 = T_1up / T_1 => T_1up = T_1*sin(39)
sin 55 = T_2up / Ts => T_2up = T_2*sin(55)
Total up tension = T_1*sin(39) + T_2*sin(55)
Newton's second law => total up tension = tension of the cable that holds the student
T_1*sin (39) + T_2*sin(55) = 711 N [equation 1]
Now find the equation from the horizontal equilibrium.
Horizontal-components fo the tension of the other two cables
cos 39 = T_1 left / T_1 => T_1 left = T_1*cos(39)
cos 55 = T_2 right / T_2=> T_2 right = T_2*cos(55)
Second Newton's Law and non movement => left-component = right component.
T_1 * cos(39) = T_2 cos(55) [equation 2]
Equation 1 and equation 2 form a systems of two equations with two variables (T_1 and T_2).
When you solve it you find:
T1 = 711 / [sin(39) + tan(55)*cos(39)] = 711 / 1.739 = 408.9 N
T_2 = cos (39)*408.9 / cos (55) = 553. 9 N
Therefore this cable will break because the tension calculated exceeds 500 N.
Make the foundation more sturdy
Answer:
The final velocity of the race car is 27.14 m/s
Explanation:
Given;
initial velocity of the race car, u = 18.5 m/s
acceleration of the race car, a = 2.47 m/s²
distance covered by the race car, s = 79.78 m
Apply the following kinematic equation to determine the final velocity of the race car.
v² = u² + 2as
v² = (18.5)² + 2(2.47)(79.78)
v² = 736.363
v = √736.363
v = 27.14 m/s
Therefore, the final velocity of the racecar is 27.14 m/s
The atomic mass of this question is 10.811
