Tipping something given its center of mass? My LAST homework problem

[Jaklynn429]

Tipping something given its center of mass? My LAST homework problem!

Homework Statement

A rectangular block of mass 3.0 kg is at rest on a table. Its height (in its current position) is 0.7 m, but because its composition varies vertically, its center of mass is at a height of 0.40 meters. Its width is 0.1 meters in both horizontal directions. At what angle does the block have to be tipped to in order to make it fall over?

Homework Equations

The Attempt at a Solution

I know that things tip when their center of mass is directly over the normal force. I think I need to make a triangle somehow but the problem has so many weird things in it that I don't know where to begin! Any help would be greatly appreciated thank you!

 

[LowlyPion]

Homework Statement

A rectangular block of mass 3.0 kg is at rest on a table. Its height (in its current position) is 0.7 m, but because its composition varies vertically, its center of mass is at a height of 0.40 meters. Its width is 0.1 meters in both horizontal directions. At what angle does the block have to be tipped to in order to make it fall over?

Homework Equations

The Attempt at a Solution

I know that things tip when their center of mass is directly over the normal force. I think I need to make a triangle somehow but the problem has so many weird things in it that I don't know where to begin! Any help would be greatly appreciated thank you!

Draw a diagram. When the center of mass is past the edge that you are tipping it about ...

So as you said it's a triangle. What is the angle with the vertical and where the center of mass is relative to the bottom edge.

It's located .4 m up and 1/2 of the width of .1 m over.

 

[baba89]

I would approach this problem by first drawing a free body diagram of the block on the table. This will help us visualize all the forces acting on the block and understand the physics behind the tipping motion.

From the diagram, we can see that the normal force (N) from the table acts upwards on the block, while the weight (mg) of the block acts downwards. The block will start to tip when the normal force is no longer acting directly under the center of mass. This can be achieved by applying a small external force (F) at the top edge of the block, causing a torque that will make the block rotate.

To find the angle at which the block will tip, we can use the formula for torque (T = r x F), where r is the distance from the point of rotation (center of mass) to the point where the external force is applied. In this case, r is equal to the height of the block (0.4 m). We also know that the torque must be equal to the weight of the block multiplied by the distance between the point of rotation and the center of mass (0.3 m).

Therefore, we can set up the equation T = r x F = mg x 0.3. Solving for the external force (F), we get F = mg x 0.3 / r. Plugging in the values, we get F = (3.0 kg)(9.8 m/s^2)(0.3 m) / 0.4 m = 6.57 N.

Now, we can use trigonometry to find the angle at which the block will tip. The external force (F) forms a right triangle with the normal force (N) and the weight (mg). We know that the tangent of the angle is equal to the opposite side (F) over the adjacent side (N). Therefore, tanθ = F / N. Plugging in the values, we get tanθ = 6.57 N / (3.0 kg)(9.8 m/s^2) = 0.225. Taking the inverse tangent, we get θ = 12.6 degrees.

Therefore, the block will tip at an angle of approximately 12.6 degrees.