Answer :
Answer:
a) a = - 0.106 m/s^2 (←)
b) T = 12215.1064 N
Explanation:
If
F₁ = 9*1350 N = 12150 N (→)
F₂ = 9*1365 N = 12285 N (←)
∑Fx = M*a = (M₁ +M₂)*a (→)
F₁ - F₂ = (M₁ +M₂)*a
→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg
→ a = - 0.106 m/s^2 (←)
(b) What is the tension in the section of rope between the teams?
If we apply ∑Fx = M*a for the team 1
F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a
⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
If we choose the team 2 we get
- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a
⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
(a) The magnitude of the acceleration of the two teams is 0.11 m/s².
(b) The tension in the section of rope between the teams is 2,715 N.
The given parameters;
- average mass of the first team, m = 68 kg
- average force of the first team, F = 1350 N
- average mass of the second team, m = 73 kg
- average force of the second team, F = 1365 N
The teams will accelerate in the direction of the greater force.
The net force on the teams is calculated as follows;
[tex]\Sigma F _{net}= 1365 - 1350 = 15 \ N[/tex]
The acceleration of the teams is calculated as follows;
[tex]a = \frac{\Sigma F_{net}}{m_1 + m_2} = \frac{15 }{68 + 73} = 0.11 \ m/s^2[/tex]
The tension in the section of rope between the teams is calculated as;
[tex]T= F_1 + F_2\\\\T= 1350 \ + \ 1365\\\\T = 2,715 \ N[/tex]
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