Question

Two children pull a third child on a tricycle by means of two ropes tied to the handlebars. The combined inertia of the child and tricycle is 34 kg . One child pulls with a constant force of 96 Ndirected to the right of the straight-ahead direction at an angle of 45∘ .

a) If the second child pulls with a force of 80 N , at what angle to the left of the straight-ahead direction must he pull to make the tricycle move straight ahead?
b) With the second child pulling at the angle you calculated in part A, what is the acceleration of the tricycle?
c) What is the work done on the tricycle by the two ropes if the children pull constantly like this while the tricycle moves 2.0 m?

Answer 1

The second child must pull at an angle of 17.5 degrees to the left of the straight-ahead direction to make the tricycle move straight ahead. The acceleration of the tricycle is 0.303 m/s^2. The work done on the tricycle by the two ropes is 346 Joules.

Step-by-step explanation:

To solve this problem, we can start by resolving the forces into their horizontal and vertical components. The force exerted by the first child can be resolved into 2 components: one along the straight-ahead direction and the other perpendicular to it. The force along the straight-ahead direction is given by:

F_parallel = F1 * cos(angle) = 96 N * cos(45) = 67.9 N

The force exerted by the second child can also be resolved into 2 components: one along the straight-ahead direction and the other perpendicular to it. Let’s call the angle at which the second child pulls F2_theta. The force along the straight-ahead direction is:

F2_parallel = F2 * cos(F2_theta)

Since we want the tricycle to move straight ahead, the sum of the forces along the straight-ahead direction should be zero:

F_parallel - F2_parallel = 0

Substituting the values we have:

67.9 N - F2 * cos(F2_theta) = 0

Given that F2 = 80 N, we can rearrange the equation to solve for F2_theta:

F2_theta = arccos(67.9 N / 80 N) = 17.5 degrees

So the second child must pull at an angle of 17.5 degrees to the left of the straight-ahead direction to make the tricycle move straight ahead.

To calculate the acceleration of the tricycle, we can use Newton’s second law:

F_parallel - F2_parallel = (m_tricycle * a)

Substituting the known values:

67.9 N - F2 * cos(F2_theta) = (34 kg * a)

Given that F2 = 80 N and F2_theta = 17.5 degrees, we can calculate the acceleration:

67.9 N - 80 N * cos(17.5) = (34 kg * a)

a = (67.9 N - 80 N * cos(17.5)) / 34 kg = 0.303 m/s^2

So the acceleration of the tricycle is 0.303 m/s^2.

To calculate the work done on the tricycle by the two ropes, we can use the formula:

Work = Force * Distance * cos(angle)

Substituting the known values:

Work = (96 N * 2.0 m * cos(45)) + (80 N * 2.0 m * cos(17.5)) = 346 J

Therefore, the work done on the tricycle by the two ropes is 346 Joules.