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 solve this problem, we need to calculate the torque exerted by each object and ensure they are balanced.
First, let's calculate the torque exerted by the plank. The torque (τ) can be calculated using the equation τ = F × d, where F is the force applied perpendicular to the distance (lever arm) from the pivot point.
The weight of the plank can be considered as acting at its center of mass, so the torque exerted by the plank (τp) is given by τp = Mg × d1, where Mg is the weight of the plank (mass × gravitational acceleration) and d1 is the distance between the center of mass and sawhorse B.
Similarly, we need to consider the torque exerted by the cat. Since the plank tips when the cat is at a certain distance from sawhorse B, we can assume that all the cat's weight is acting at that distance. Thus, the torque exerted by the cat (τc) is given by τc = Mcat × g × d2, where Mcat is the mass of the cat, g is the gravitational acceleration, and d2 is the distance of the cat from sawhorse B.
Since the plank just begins to tip when the torques are balanced, we can set up the equation τp = τc.
Mg × d1 = Mcat × g × d2
Dividing both sides by g, we get:
M × d1 = Mcat × d2
Now, we can plug in the given values:
7.00 kg × 0.850 m = 3.6 kg × d2
Rearranging the equation, we find:
d2 = (7.00 kg × 0.850 m) / 3.6 kg
d2 ≈ 1.648 m
Therefore, the cat can walk approximately 1.648 meters to the right of sawhorse B before the plank begins to tip.