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Chapter 4: Problem 57

4.57 Your refrigerator has a mass of \(112.2 \mathrm{~kg}\), including the food in it. It is standing in the middle of your kitchen, and you need to move it. The coefficients of static and kinetic friction between the fridge and the tile floor are 0.460 and 0.370 , respectively. What is the magnitude of the force of friction acting on the fridge, if you push against it horizontally with a force of each magnitude? a) \(300 \mathrm{~N}\) b) \(500 \mathrm{~N}\) c) \(700 \mathrm{~N}\)

Short Answer

Expert verified
Answer: The magnitudes of the force of friction for each applied force are: a) 300 N b) 500 N c) 406.69 N

Step by step solution

01

Calculate the Normal Force

To calculate the normal force, multiply the mass of the refrigerator by the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)). \(N = m \times g = 112.2 \mathrm{~kg} \times 9.8 \mathrm{m/s^2} = 1099.16 \mathrm{~N}\)
02

Calculate the Maximum Force of Static Friction

Use the static friction coefficient (0.460) and the normal force calculated in Step 1 to find the maximum force of static friction. \(F_{s_{max}} = \mu_s * N = 0.460 * 1099.16 \mathrm{~N} = 505.21 \mathrm{~N}\) Now let's calculate the kinetic friction (\(F_k\)) for when the refrigerator is moving: \(F_k = \mu_k * N\) Where \(\mu_k\) is the kinetic friction coefficient.
03

Calculate the Kinetic Friction

Use the kinetic friction coefficient (0.370) and the normal force calculated in Step 1 to find the kinetic friction. \(F_k = \mu_k * N = 0.370 \times 1099.16 \mathrm{~N} = 406.69 \mathrm{~N}\) Now that we have the maximum force of static friction and the kinetic friction, we can find the force of friction for each applied force magnitude. Remember to check if the applied force is greater than the maximum force of static friction (if yes, the refrigerator will move and we should use \(F_k\), otherwise we use applied force when the refrigerator doesn't move).
04

Calculate the Force of Friction for Each Magnitude

a) For an applied force of \(300 \mathrm{~N}\), Since \(300 \mathrm{~N} < F_{s_{max}}\), the refrigerator doesn't move, so the force of friction is equal to the applied force. Force of friction = \(300 \mathrm{~N}\) b) For an applied force of \(500 \mathrm{~N}\), Since \(500 \mathrm{~N} < F_{s_{max}}\), the refrigerator doesn't move, so the force of friction is equal to the applied force. Force of friction = \(500 \mathrm{~N}\) c) For an applied force of \(700 \mathrm{~N}\), Since \(700 \mathrm{~N} > F_{s_{max}}\), the refrigerator moves, so the force of friction is equal to the kinetic friction. Force of friction = \(406.69 \mathrm{~N}\) Hence, the magnitudes of the force of friction for each applied force are: a) \(300 \mathrm{~N}\) b) \(500 \mathrm{~N}\) c) \(406.69 \mathrm{~N}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction
Static friction is the force that keeps an object at rest when you apply a small force. It acts in the opposite direction to the applied force, preventing the object from sliding.
This type of friction is always equal to or less than its maximum value, and it adjusts itself to counteract any force applied until the limit is reached.
The maximum static friction can be calculated as follows:\[F_{s_{max}} = \mu_s \times N\]Where:
  • \(\mu_s\) is the coefficient of static friction, a value that represents the frictional properties of the interfaces in contact (in our example, it is 0.460).
  • \(N\) is the normal force, essentially the weight of the object acting perpendicular to the surface.
If the force you apply is less than \(F_{s_{max}}\), the object doesn鈥檛 move, and the force of friction equals the applied force. Unlike kinetic friction, static friction increases with the applied force until it reaches its peak value.
Kinetic Friction
Once an object starts to move, static friction is no longer acting. Instead, kinetic friction comes into play. While static friction prevents motion, kinetic friction opposes motion that is already happening.
It is generally lower than the maximum static friction, which is why it's easier to keep an object moving than to start moving it from rest.
The force of kinetic friction \(F_k\) can be determined by:\[F_k = \mu_k \times N\]Where:
  • \(\mu_k\) is the coefficient of kinetic friction, the ratio that reduces the normal force when an object is sliding across a surface (in our case, \(\mu_k = 0.370\)).
  • \(N\), again, is the normal force. Kinetic friction does not vary with speed, as long as the object is in motion.
With kinetic friction, once you push past the maximum static friction and the object begins to move, the frictional force becomes constant and is equal to the value calculated using the kinetic friction equation.
Normal Force
The normal force plays a critical role in calculating both static and kinetic frictions. It is the support force exerted by a surface perpendicular to the force of gravity acting on the object.
The normal force is directly tied to the object鈥檚 weight and is calculated in cases where the surface is flat as:\[N = m \times g\]Where:
  • \(m\) is the mass of the object (for the refrigerator, it is \(112.2\,\mathrm{kg}\)).
  • \(g\) is the acceleration due to gravity, approximately \(9.8\,\mathrm{m/s^2}\).
In the context of this exercise, the normal force remains the same for all scenarios since neither the surface nor the mass of the refrigerator changes.

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Most popular questions from this chapter

4.71 A block of mass \(20.0 \mathrm{~kg}\) supported by a vertical massless cable is initially at rest. The block is then pulled upward with a constant acceleration of \(2.32 \mathrm{~m} / \mathrm{s}^{2}\). a) What is the tension in the cable? b) What is the net force acting on the mass? c) What is the speed of the block after it has traveled \(2.00 \mathrm{m?}\)

A crane of mass \(M=1.00 \cdot 10^{4} \mathrm{~kg}\) lifts a wrecking ball of mass \(m=1200 .\) kg directly upward. a) Find the magnitude of the normal force exerted on the crane by the ground while the wrecking ball is moving upward at a constant speed of \(v=1.00 \mathrm{~m} / \mathrm{s}\). b) Find the magnitude of the normal force if the wrecking ball's upward motion slows at a constant rate from its initial speed \(v=1.00 \mathrm{~m} / \mathrm{s}\) to a stop over a distance \(D=0.250 \mathrm{~m}\)

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.

A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of \(5.832 \mathrm{~m} / \mathrm{s}\) after sliding a distance of \(2.29 \mathrm{~m},\) what is the angle of inclination of the plane with respect to the horizontal?

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?
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