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jenyasd209 [6]
3 years ago
15

Nick, a 60 kg physics student, wants to go bungee jumping, but doesn't have a bungee cord. He finds a 15 m long, strong spring (

with a spring constant of 60 N/m) in his garage, which he decides should work just as well. He takes the spring to a nearby bridge, attaches one end to the bridge and the other to his leg and jumps off. After bouncing around violently for a while, he finally comes to rest. How far below the bridge is he hanging
Physics
1 answer:
vitfil [10]3 years ago
4 0

Answer:

h = 24.81 m

Explanation:

Given:-

- The mass of the student, m = 60 kg

- The length of the spring, L = 15 m

- The spring constant, k = 60 N/m

Find:-

How far below the bridge is he hanging

Solution:-

- First realize that after the student attempts a bungee jump he oscillates violently ( dynamic motion ). After some time all the kinetic energy has been converted to Elastic and gravitational potential energy student is (stationary) and hanging down on one end of the spring.

- We will apply equilibrium condition on the student. We see that there are two forces acting on the student. The weight (W) of the student acting downward is in combat with the spring restoring force (Fs) acting upwards.

- Apply equilibrium condition in vertical direction:

                               Fs - W = 0

                               Fs = W

- The weight and spring force can be expressed as:

                               k*x = m*g

Where,     g : Gravitational acceleration constant = 9.81 m/s^2

                x : The extension of the spring from original position

- Solve for the extension (x) of the spring for this condition.

                              x = m*g / k

- Plug in the values and evaluate:

                              x = (60 kg)*(9.81 m/s^2) / (60 N/m)

                              x = 9.81 m

- The spring extends for about 9.81 m from its original length. So the distance (h) from edge of the bridge would be:

                              h = L + x

                              h = 15 + 9.81

                              h = 24.81 m

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Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

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After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average
lara [203]

Answer:

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Explanation:

Given:

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- The total number of goats n_g = 3

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- Compute the force due to all the weights on the taptap:

                                F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

                                F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81  

                                F_weight = 19109.88 N

- The restoring force of a spring is given by:

                                F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

                                 F_weight = F_spring

                                 19109.88 = k*x

                                 x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

                                 x = 19109.88 / 160,000

                                 x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

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