Example 11-9 depicts the following scenario. A cat walks along a plank with mass M= 7.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a distance d1= 0.850 m to the left of sawhorse B. When the cat is a distance d2= 1.11 m to the right of sawhorse B, the plank just begins to tip.
If the cat has a mass of 3.6 kg, how far to the right of sawhorse B can it walk before the plank begins to tip?
To answer this question, we need to analyze the torque acting on the system. The plank will tip if the torque caused by the weight of the cat is greater than the torque caused by the weight of the plank and the sawhorses.
The torque caused by a force is defined as the product of the magnitude of the force and the perpendicular distance from the point of rotation to the line of action of the force. In this case, the point of rotation is the left end of the plank.
Let's denote the distance to the center of mass of the plank as d1 = 0.850 m and the distance to the right end of the plank (where the cat would be) as d2.
The torque caused by the weight of the cat is given by: T_cat = (m_cat * g * d2), where m_cat is the mass of the cat and g is the acceleration due to gravity.
The torque caused by the weight of the plank and the sawhorses is given by: T_plank = (m_plank * g * d1), where m_plank is the mass of the plank (7.00 kg in this case).
The plank will begin to tip when T_cat becomes equal to T_plank. So, we can set up the equation:
(m_cat * g * d2) = (m_plank * g * d1)
Plugging in the given values, we have:
(3.6 kg * 9.8 m/s^2 * d2) = (7.00 kg * 9.8 m/s^2 * 0.850 m)
Simplifying the equation, we find:
3.6 * d2 = 7.00 * 0.850
Dividing both sides by 3.6:
d2 = (7.00 * 0.850) / 3.6
Calculating this expression, we get:
d2 ≈ 1.6483 m
Therefore, the cat can walk to a distance of approximately 1.6483 meters to the right of sawhorse B before the plank begins to tip.
To find the distance to the right of sawhorse B where the plank just begins to tip, we can use the principle of rotational equilibrium.
The torque about sawhorse B caused by the mass of the cat and the plank must be equal to zero. The torque is calculated as the force multiplied by the perpendicular distance to the pivot point.
Let's start by finding the torque caused by the mass of the cat. The weight of the cat can be calculated as the mass multiplied by the acceleration due to gravity (9.8 m/s^2).
Torque due to the cat = weight of the cat * perpendicular distance
= (mass of the cat * acceleration due to gravity) * d2
Next, we need to calculate the torque caused by the plank. The weight of the plank can be calculated using the same formula.
Torque due to the plank = weight of the plank * perpendicular distance
= (mass of the plank * acceleration due to gravity) * d1
The sum of these two torques must be zero for the plank to be in rotational equilibrium.
Torque due to the cat + Torque due to the plank = 0
[(mass of the cat * acceleration due to gravity) * d2] + [(mass of the plank * acceleration due to gravity) * d1] = 0
We can rearrange this equation to solve for the distance d2:
d2 = -[(mass of the plank * acceleration due to gravity) * d1] / (mass of the cat * acceleration due to gravity)
Substituting the given values:
mass of the plank = 7.00 kg
d1 = 0.850 m
mass of the cat = 3.6 kg
d2 = -[(7.00 kg * 9.8 m/s^2) * 0.850 m] / (3.6 kg * 9.8 m/s^2)
After calculating this expression, we find that d2 is approximately -1.051 m.
Therefore, the cat can walk a maximum distance of approximately 1.051 m to the right of sawhorse B before the plank begins to tip.