A 49-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity does not lie halfway between the two ends. The system is balanced by a support that is 0.30 m from the left end of the plank. How far to the right of the support is the plank’s center of gravity?
21 (x) = 49(.3)
x = .7 m
To find the center of gravity of the plank, we need to consider the torques acting on the system.
First, let's calculate the torque exerted by the crate. Torque is given by the product of force and the perpendicular distance from the pivot point (support) to the line of action of the force.
Considering the crate as a point mass, the torque exerted by the crate is:
Torque(crate) = force(crate) * distance(crate from support)
= 49 N * 0.30 m
Next, let's calculate the torque exerted by the plank itself. Since the plank is not uniform, its center of gravity does not lie halfway between the two ends. Let's assume the distance from the left end of the plank to its center of gravity is x meters.
The total length of the plank is L = distance(support to right end) + distance(support to left end)
= x + (L - x)
= L
The torque exerted by the plank can be calculated using its weight and the distance from the center of gravity to the support.
Torque(plank) = force(plank) * distance(plank's center of gravity from support)
= 21 N * (x - 0.30 m)
Since the system is balanced, the sum of the torques exerted by the crate and the plank will be zero.
Torque(crate) + Torque(plank) = 0
Substituting the values, we get:
49 N * 0.30 m + 21 N * (x - 0.30 m) = 0
Now, we can solve this equation to find the value of x, which represents the distance from the left end of the plank to the center of gravity.
147 N + 21 N * x - 6.3 N = 0
21 N * x = 6.3 N - 147 N
21 N * x = -140.7 N
x = -140.7 N / 21 N
x ≈ -6.71 m
Since distance cannot be negative, it means the center of gravity of the plank is approximately 6.71 m to the right of the support.