You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0.73, what is the mass of the heaviest box you will be able to move?
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Well, well, well, looks like we've got ourselves a classic physics problem here! Let's see if we can solve it with a touch of humor.
First things first, we need to find the vertical component of the force you're applying. So, we take the sine of the angle (25° below the horizontal) and multiply it by the force of 750 N. Let's call this component "Fv" because it's so vertical, it could audition for a Cirque du Soleil act.
Fv = 750 N * sin(25°)
Next, we need to figure out the amount of friction holding this box back. We use the coefficient of static friction, which is like the box's laziness level. The friction force is then equal to the coefficient of static friction multiplied by the weight of the box (mg).
Friction force (Ff) = coefficient of static friction * weight of the box
But we're not done, my friend! The weight of the box is just the mass of the box (m) multiplied by the acceleration due to gravity (g). Don't worry, gravity won't judge you, it's too busy keeping us grounded.
Weight of the box (mg) = mass of the box (m) * acceleration due to gravity (g)
Now, plug that weight back into the equation for friction force:
Ff = coefficient of static friction * (mass of the box * acceleration due to gravity)
But hold on a second! We need to know the mass of the heaviest box you can move. And to find that, we need to know the maximum friction force (Ff_max) that can be applied before the box overcomes its laziness and starts moving. Luckily, we can calculate that too.
Ff_max = coefficient of static friction * (mass of the box * acceleration due to gravity)
Now, here's where the magic happens. If you're able to move the box, that means the vertical force you're applying (Fv) is larger than the maximum friction force (Ff_max). So, we set up an inequality:
Fv > Ff_max
Now, we can substitute in the formulas we derived earlier:
750 N * sin(25°) > 0.73 * (m * 9.8 m/s²)
Solve that inequality for (m), and you'll get the maximum mass you can lift! Hooray!
Just remember, even if the box is too heavy to move, you can always just crack a few jokes to lighten the mood. Keep a smile on your face, my friend!
To solve this problem, we need to use the concept of equilibrium and the laws of static friction.
Step 1: Draw a Free Body Diagram (FBD) to visualize the forces acting on the box. In this case, we have the weight of the box acting downward (mg), the normal force (N) acting perpendicular to the surface, and the static friction force (fs) acting parallel to the surface.
Step 2: Break down the given force into its horizontal and vertical components. The force you apply can be split into two components: a horizontal component (Fh) and a vertical component (Fv). The horizontal component (Fh) can be determined using the angle of 25°:
Fh = 750 N * cos(25°)
Fh ≈ 679.43 N
Step 3: Determine the maximum static friction force (fs max) using the formula:
fs max = μs * N
Where μs is the coefficient of static friction and N is the normal force. Since the box is on a horizontal surface, the normal force (N) is equal to the weight of the box (mg).
Step 4: Calculate the normal force (N). Since the box is on a flat horizontal surface, the normal force (N) is equal to the weight of the box (mg).
Step 5: Calculate the maximum mass (m max) of the box that you can move by setting the static friction force (fs) equal to the maximum static friction force (fs max):
fs = fs max
μs * N = mg
μs * m * g = m * g
μs = m
Step 6: Substitute μs with the coefficient of static friction (0.73):
0.73 * m = m
Step 7: Solve for the mass (m):
0.73 * m = m
0.73m - m = 0
0.27m = 0
m = 0 / 0.27
m = 0
Therefore, the mass of the heaviest box you will be able to move is zero. This implies that you will not be able to move any box, regardless of its weight, with a downward force of 750 N at an angle of 25° below the horizontal.
54
normal force on floor = 750 sin 25 + 9.81 m
friction force =.73(750 sin 25 + 9.81 m)
so
750cos25 = .73(750 sin 25 + 9.81 m)
solve for m