Calculating Force to Tip a Box Dragged Across Floor

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In summary, the problem involves dragging a box of unknown mass (M) across the floor with a rope at a height (h) above the floor. The coefficient of friction is 0.35 and the box is 0.50m high and 0.25m wide. The goal is to find the force (F) that will just tip the box over. The solution requires the use of simultaneous equations involving sum of torques and forces, including the torque from the pseudo inertial force acting at the center of mass. However, the problem does not provide all the necessary information, such as the mass of the box and the inertial force at the center of mass.
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Homework Statement


A box is being dragged across the floor with a rope that is h distance above the floor. The coefficient of friction is 0.35. The box is 0.50m high and 0.25m wide. Find the force that just tips the box.


Homework Equations


Ff=uMg=3.43m/s2M
Fg=Mg=9.80m/s2M
Well,...that's about it.


The Attempt at a Solution


I don't understand the problem enough to draw a force/torque diagram! But I do know that h and M are going to be unknown variables.
 
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  • #2
There are two forces of interest.

The weight acting through the center of mass.
The force acting at height h.

So when the force moment, F * h is greater than m*g*width/2 acting down, it should tip about the leading corner right?
 
  • #3
The problem should identify the mass of the box as M. Then M and h are given variables. F (the applied tipping force) is the unknown variable. I don't know why the problem left out the "M". But note that since the block is sliding and has overcome static friction without tipping, it must be accelerating. It can't be moving at constant speed at the 'tip over' point. The solution requires the use of simultaneous eqations involving sum of torques =0 , including the torque from the pseudo inertial force acting at the c.g (which is also not given) , and sum of forces in x direction, including the pseudo inertial force acting at the c.g , = 0. Poorly worded problem when you have to make your own assumptions.
 
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1. How do you calculate the force needed to tip a box dragged across a floor?

To calculate the force needed to tip a box dragged across a floor, you need to consider the weight of the box, the coefficient of friction between the box and the floor, and the angle at which the box is being dragged. The formula for calculating this force is: Force = (Weight of the box x Coefficient of friction) / sin(angle).

2. What is the coefficient of friction and how does it affect the force required to tip a box?

The coefficient of friction is a measure of the resistance between two surfaces in contact. It is a unitless value that ranges from 0 (no friction) to 1 (maximum friction). The higher the coefficient of friction, the more force is required to move the box and thus, the harder it is to tip the box.

3. How does the weight of the box impact the force needed to tip it?

The weight of the box directly affects the force needed to tip it. The heavier the box, the more force is needed to overcome its weight and the harder it is to tip. Therefore, a lighter box will require less force to tip compared to a heavier box.

4. What angle should the box be dragged at to require the minimum force to tip it?

The minimum force required to tip a box dragged across a floor can be achieved when the angle at which the box is being dragged is 90 degrees (perpendicular to the floor). This is because at this angle, the force needed to overcome the weight of the box is the lowest.

5. Can the force needed to tip a box vary depending on the type of floor it is being dragged across?

Yes, the type of floor can affect the coefficient of friction between the box and the floor, thereby affecting the force needed to tip the box. For example, a smoother floor may have a lower coefficient of friction and thus, require less force to tip the box compared to a rougher floor with a higher coefficient of friction.

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