A school yard teeter totter with a total length of 5.46 m and a mass of 35.9 kg is pivoted at its center. A 17.7 kg child sits on one end of the teeter totter. Where should a parent push with a force of 272 N in order to hold the teeter totter level?
sum moments about the center.
5.46/2 * 17.7g-272N*d=0 the sum of moments about any point equals zero. d is the distance away from the child, on the opposide side.
Since the board is balanced at the center, the weight of the board does not contribute a moment.
You must balance the moment (force x distance) applied by the parent with the moment due to the weight of the child.
272*X = (17.7 kg)*g*(2.73 m)
Solve for X in meters
Note that the child's mass has to be multiplied by g to get the weight force in Newtons.
To hold the teeter totter level, the torques on both sides of the pivot point should be equal.
Torque is calculated by multiplying the force applied by the distance from the pivot point.
Let x be the distance from the pivot point to where the parent should push.
On one side with the child, the torque is given by:
Torque child = weight of child * distance from pivot to child
= 17.7 kg * 9.8 m/s^2 * x
On the other side with the parent, the torque is given by:
Torque parent = force applied by parent * distance from pivot to parent
= 272 N * (5.46 m - x)
Setting up the equation with equal torques:
Torque child = Torque parent
17.7 kg * 9.8 m/s^2 * x = 272 N * (5.46 m - x)
Simplifying the equation:
173.46 * x = 272 * (5.46 - x)
173.46x = 1483.92 - 272x
Combining like terms:
173.46x + 272x = 1483.92
445.46x = 1483.92
Solving for x:
x = 1483.92 / 445.46
x ≈ 3.33 m
Therefore, the parent should push at a distance of approximately 3.33 meters from the pivot point in order to hold the teeter totter level.
To find where the parent should push with a force of 272 N in order to hold the teeter totter level, we can use the principle of torque equilibrium.
The torque exerted by the parent pushing on the teeter totter is given by the formula:
τ = r * F
Where τ is the torque, r is the perpendicular distance from the pivot point to the line of action of the force, and F is the magnitude of the force.
To hold the teeter totter level, the sum of the torques on either side of the pivot point should be zero.
In this case, we know that the mass and distance from the pivot point for the teeter totter and the child sitting on it. Let's call the distance from the pivot point to the child's end of the teeter totter as x.
The torque exerted by the child sitting on the teeter totter is given by:
τ_child = x * F_child
Where F_child is the weight of the child, which can be calculated by multiplying the mass of the child by the acceleration due to gravity (F_child = m_child * g).
Similarly, the torque exerted by the parent pushing on the teeter totter is given by:
τ_parent = (5.46 - x) * F_parent
Where F_parent is the force exerted by the parent, which is 272 N in this case.
Since the teeter totter is held level, the sum of the torques on either side of the pivot point should be zero:
τ_child + τ_parent = 0
Substituting the equations for torque, we get:
x * F_child + (5.46 - x) * F_parent = 0
Plugging in the given values:
x * (m_child * g) + (5.46 - x) * 272 = 0
Now we can solve this equation to find the value of x. Once we have the value of x, we can determine where the parent should push with a force of 272 N by calculating the distance from the pivot point to the parent's end of the teeter totter (which would be 5.46 - x).