A thin 2kg box rests on a 6kg board 50cm long with 20cm hanging off the edge of a table. How far can the center of the box be from the end of the table before the board begins to tilt?
According to the above reference the answer would be 45cm yes? - which doesn't make sense because there is only 20cm of board from the end of the table. And i'm not exactly sure where the 25cm came from? Can someone please expand on this answer?
If you assume the cg of the board is at the center, then summing moments from the end of the board on the table.
L= length of box from edge of table.
2kg*g*L+ 6kg*g*25cm-(8kg*g*30cm)=0
solve for L. NOtice I put the force of the table at the edge, which is true if the board is fixing to rotate and the box falls.
To determine how far the center of the box can be from the end of the table before the board begins to tilt, we need to find the point at which the torque exerted by the box is equal to the torque exerted by the board.
Torque is calculated by multiplying the force applied by the distance from the pivot point. In this case, the pivot point is the edge of the table.
First, let's calculate the torque exerted by the box. The weight of the box is given as 2kg, so the force it exerts is equal to its weight times the acceleration due to gravity, which is 9.8 m/s². Therefore, the force exerted by the box is:
Force = 2kg × 9.8 m/s² = 19.6 N
Since the center of the box is located at half of its length (L/2), the distance from the pivot point (edge of the table) to the center of the box is:
Distance_box = L/2 = 50cm/2 = 25cm = 0.25m
Now, let's calculate the torque exerted by the box:
Torque_box = Force × Distance_box = 19.6 N × 0.25 m = 4.9 N·m
Next, let's calculate the torque exerted by the board. The weight of the board is given as 6kg, so the force it exerts is equal to its weight times the acceleration due to gravity. The force exerted by the board is:
Force = 6kg × 9.8 m/s² = 58.8 N
Since the board is hanging off the table by 20cm, the distance from the pivot point (edge of the table) to the end of the board is:
Distance_board = 20cm = 0.2m
Now, let's calculate the torque exerted by the board:
Torque_board = Force × Distance_board = 58.8 N × 0.2 m = 11.76 N·m
In order for the board to remain balanced and not tilt, the torque exerted by the box must be equal to the torque exerted by the board:
Torque_box = Torque_board
4.9 N·m = 11.76 N·m
We can rearrange this equation to find the distance from the pivot point to the center of the box:
Distance_box = Torque_board / Force = 11.76 N·m / 19.6 N = 0.6 m
Therefore, the center of the box can be up to 0.6 meters (or 60cm) from the end of the table before the board begins to tilt.
To determine how far the center of the box can be from the end of the table before the board begins to tilt, we need to consider the center of mass and the torques acting on the system.
The center of mass for the combined box and board system can be calculated based on the masses and the positions of their centers of mass. Since the box is thin and has a uniform density, its center of mass will be at the geometric center, which is halfway along its length. So, the center of mass for the box is at the 25cm mark.
For the board, since it is symmetric about its center, the center of mass will also be at the 25cm mark.
To prevent the board from tilting, the torque exerted by the weight of the box must be balanced by the torque exerted by the weight of the board.
The torque exerted by an object can be calculated as the product of the weight (mg) and the perpendicular distance (d) from the object's center of mass to the pivot point. In this case, the pivot point is the edge of the table.
Let's assume that the distance from the center of mass of the box to the end of the table is x cm. The distance from the center of mass of the board to the end of the table is (50 - x) cm.
The torque exerted by the weight of the box is given by (2 kg * 9.8 m/s²) * (x cm / 100 cm).
The torque exerted by the weight of the board is given by (6 kg * 9.8 m/s²) * ((50 - x) cm / 100 cm).
For the board not to tilt, the torques exerted by the weight of the box and the weight of the board must be equal:
(2 kg * 9.8 m/s²) * (x cm / 100 cm) = (6 kg * 9.8 m/s²) * ((50 - x) cm / 100 cm).
Simplifying the equation gives:
0.02x = 0.294(50 - x).
Solving for x:
0.02x = 0.294(50) - 0.294x.
0.02x + 0.294x = 14.7.
0.314x = 14.7.
x ≈ 46.7 cm.
Therefore, the center of the box can be approximately 46.7 cm from the end of the table before the board starts to tilt.