The cue is an added load on the same feet, followed by a change in pressure, or pounds of force per square inch. Subtract the pressures to isolate the box’s effect, then divide the box’s weight equally among the feet. Use those two changes to find one foot’s contact area. The tempting trap is to give the added pounds on a foot as its area.
Hints
- Hint 1
The workbench’s own weight hasn’t changed, and it stands on the same feet. A change in pressure comes from the box alone. Which two pressure readings should you subtract?
- Hint 2
Because the box’s weight is shared equally among the feet, each foot supports one-fourth of the added weight. That gives pounds on one foot, not its area. How many pounds does the box add to each foot?
- Hint 3
Pressure means pounds divided by square inches. Use the added pounds on one foot and its pressure increase to write an equation for that foot’s area.
Step-by-step
Use the change in pressure
Step 1Find the pressure added by the box
The workbench’s own weight hasn’t changed, so subtract the two pressures to isolate the box’s contribution:
- Step 2
Find the added weight on one foot
The box weighs pounds, shared equally among feet. Divide to find the added force, or weight pressing down, on each foot:
- Step 3
Write an equation for one foot’s area
Let be one foot’s contact area in square inches. Pressure is force divided by area. The box adds pounds to that foot and raises its pressure by pounds per square inch, so write:
Use the increase, not : the full pressure includes the workbench’s weight.
- Step 4
Solve for the contact area
Type into Desmos. The tells Desmos to solve for the area rather than graph , and tells it to match the two sides. Under PARAMETERS, Desmos shows . So the bottom of each foot touches square inches of floor. Choice B.