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(o) A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. 6-17). The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N},(\mathrm{~b}) 10 \mathrm{~N}\), and (c) \(12 \mathrm{~N}\).

Short Answer

Expert verified
Frictional force: (a) 6.0 N, (b) 3.625 N, (c) 3.125 N.

Step by step solution

01

Analyze forces and identify components

Consider the forces acting on the block: the weight due to gravity, the horizontal force \( \vec{F} = 6.0 \mathrm{~N} \), the vertical force \( \vec{P} \), and the normal force. The frictional force will oppose the applied horizontal force, and its magnitude will be determined by the normal force, \( N \).
02

Calculate the normal force (N) for each scenario

The normal force \( N \) is the force exerted by the surface, and it's affected by any vertical components. It is given by \( N = mg - P \) where \( m = 2.5 \mathrm{~kg} \) and \( g = 9.8 \mathrm{~m/s^2} \). Compute \( N \) for \( P = 8.0 \mathrm{~N}, 10 \mathrm{~N}, \) and \( 12 \mathrm{~N} \).
03

Calculate the maximum static friction force for each scenario

The maximum static friction is \( f_{s, \, \text{max}} = \mu_s N \). Use \( \mu_s = 0.40 \) and the calculated \( N \) from Step 2 to find \( f_{s, \, \text{max}} \) when \( P = 8.0 \mathrm{~N}, 10 \mathrm{~N}, \) and \( 12 \mathrm{~N} \).
04

Compare applied horizontal force with maximum static friction

If the applied horizontal force \( \vec{F} = 6.0 \mathrm{~N} \) is less than or equal to \( f_{s, \, \text{max}} \), then the block does not move, and frictional force \( f = \vec{F} \). If \( \vec{F} > f_{s, \, \text{max}} \), kinetic friction formula \( f_k = \mu_k N \) with \( \mu_k = 0.25 \) is used.
05

Calculate frictional force for each value of \( P \)

Compute the frictional force:- For \( P = 8.0 \mathrm{~N} \): \( N = 2.5 \times 9.8 - 8 = 16.5 \mathrm{~N} \). \( f_{s, \, \text{max}} = 0.40 \times 16.5 \mathrm{~N} = 6.6 \mathrm{~N} \). Since \( 6.0 \mathrm{~N} < 6.6 \mathrm{~N} \), friction is \( 6.0 \mathrm{~N} \).- For \( P = 10.0 \mathrm{~N} \): \( N = 2.5 \times 9.8 - 10 = 14.5 \mathrm{~N} \). \( f_{s, \, \text{max}} = 0.40 \times 14.5 = 5.8 \mathrm{~N} \). Since \( 6.0 \mathrm{~N} > 5.8 \mathrm{~N} \), use kinetic force: \( f = 0.25 \times 14.5 \mathrm{~N} = 3.625\mathrm{~N} \). - For \( P = 12.0 \mathrm{~N} \): \( N = 2.5 \times 9.8 - 12 = 12.5 \mathrm{~N} \). \( f_{s, \, \text{max}} = 0.40 \times 12.5 = 5.0 \mathrm{~N} \). Since \( 6.0 \mathrm{~N} > 5.0 \mathrm{~N} \), use kinetic force: \( f = 0.25 \times 12.5 \mathrm{~N} = 3.125 \mathrm{~N} \).

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Key Concepts

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

Static Friction Coefficient
Static friction acts as a resisting force between two surfaces that are not in motion relative to each other. Its primary role is to hold the object in place until enough force is applied to overcome it. This property is quantified by the static friction coefficient, denoted as \( \mu_s \). This coefficient is a unitless number representing how resistant a surface is to initial motion. The larger \( \mu_s \), the more force is needed to start movement.

To determine the maximum static friction force, we use the formula:
  • \( f_{s, \text{max}} = \mu_s \times N \)
Here, \( N \) is the normal force, which we'll explore further. A key point is that the static friction force can adjust up to its maximum value, meaning that it increases as needed to match the applied force until it reaches \( f_{s, \text{max}} \).

In practice, once an object starts to move, static friction gives way to kinetic friction. Thus, understanding this coefficient is crucial for predicting when movement will begin.
Kinetic Friction Coefficient
Once an object overcomes static friction and starts moving, kinetic friction takes over. This is where the kinetic friction coefficient, \( \mu_k \), becomes important. Just like the static friction coefficient, \( \mu_k \) is a unitless factor indicating the level of friction between moving surfaces.

Generally, \( \mu_k \) is less than \( \mu_s \), reflecting the decrease in force required to keep an object moving compared to getting it started. The formula to calculate the kinetic friction force is:
  • \( f_k = \mu_k \times N \)
The kinetic frictional force remains constant as long as the object moves at a steady pace. If an object is accelerating or decelerating, other forces are involved, but the pressure exerted by kinetic friction remains calculated by this straightforward formula.

In situations with varying applied forces, like in the exercise example, it's essential to compare \( \mu_s \) and \( \mu_k \) to determine when the block transitions from static to kinetic friction.
Normal Force
The normal force \( N \) plays a critical role in the calculation of both static and kinetic friction. It acts perpendicular to the surfaces in contact and essentially supports the weight of the resting object.

Calculating the normal force often involves considering all vertical forces in a given problem. For a flat surface, \( N \) can be derived from the object's weight and any additional vertical forces applied, such as in the problem where a different force \( \vec{P} \) is considered. The formula used in such a scenario would be:
  • \( N = mg - P \)
Where \( m \) is the mass of the object, \( g \) is the gravitational acceleration, and \( P \) is an external vertical force. Negative contributions to \( N \) (such as \( P \)) occur when a force acts upwards, opposing the weight.

By accurately determining \( N \), we can ensure precise calculations for static and kinetic friction, which are crucial for understanding an object's motion under various forces.

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Most popular questions from this chapter

ssM A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68\), and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) 222 \(\mathrm{N}\), (b) \(334 \mathrm{~N},(\mathrm{c}) 445 \mathrm{~N}\), and \((\mathrm{d}) 556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

A child places a picnic basket on the outer rim of a merry-goround that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

minz A sling-thrower puts a stone \((0.250 \mathrm{~kg})\) in the sling's pouch \((0.010 \mathrm{~kg})\) and then begins to make the stone and pouch move in a vertical circle of radius \(0.650 \mathrm{~m}\). The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is \(33.0 \mathrm{~N}\) or more. Suppose the slingthrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?

SSM A warehouse worker exerts a constant horizontal force of magnitude \(85 \mathrm{~N}\) on a \(40 \mathrm{~kg}\) box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of \(1.4 \mathrm{~m}\), its speed is \(1.0 \mathrm{~m} / \mathrm{s}\). What is the coefficient of kinetic friction between the box and the fioor?

SSM A \(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu_{k}=0.35\), what is the magnitude of the initial acceleration of the crate?

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