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Chapter 6: Problem 11

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?

Short Answer

Expert verified
The minimum force required is approximately 320.33 N, and the crate's initial acceleration is about 0.57 m/s虏.

Step by step solution

01

Identifying Forces Acting on the Crate

First, identify the forces acting on the crate. These include the gravitational force (weight of the crate), the normal force, the force applied via the rope, and the frictional force. The weight, \(W\), is given as \(mg\) where \(m = 68 \text{ kg}\) and \(g = 9.8 \text{ m/s}^2\), so \(W = 68 \times 9.8 = 666.4 \text{ N}\).
02

Resolve the Rope Tension into Components

The force exerted by the rope has components: a horizontal component \(F \cos(15^{\circ})\) and a vertical component \(F \sin(15^{\circ})\). These components will interact with the frictional forces and normal forces.
03

Calculate Normal Force

The normal force, \(N\), can be found from the balance of vertical forces: \(N + F \sin(15^{\circ}) = mg\). Therefore: \(N = mg - F \sin(15^{\circ})\).
04

Static Friction and Minimum Force to Start Movement

The maximum static friction is \(f_s = \mu_s N = 0.50(mg - F \sin(15^{\circ}))\). The crate starts to move when the horizontal component of the tension \(F \cos(15^{\circ})\) is equal to \(f_s\). Equating these gives: \[ F \cos(15^{\circ}) = 0.50(mg - F \sin(15^{\circ})) \] solving this will give you the minimum force \(F\).
05

Solve for Maximum Static Force

Re-arrange and solve the equations for \(F\): \[ F (\cos(15^{\circ}) + 0.50 \sin(15^{\circ})) = 0.50 \times 666.4 \] \[ F = \frac{0.50 \times 666.4}{\cos(15^{\circ}) + 0.50 \sin(15^{\circ})} \]
06

Calculate Initial Acceleration with Kinetic Friction

Once the crate is moving, switch to kinetic friction: \(f_k = \mu_k N = 0.35(mg - F \sin(15^{\circ}))\). The net horizontal force is \(F \cos(15^{\circ}) - f_k\) which equals \(ma\). Solve \(a = \frac{F \cos(15^{\circ}) - 0.35(mg - F \sin(15^{\circ}))}{m}\) to find acceleration.
07

Solve for Acceleration

Plug the values into the equation from Step 6 and solve for the acceleration \(a\): \[ a = \frac{F \cos(15^{\circ}) - 0.35(666.4 - F \sin(15^{\circ}))}{68} \] This will give you the magnitude of the initial acceleration of the crate.

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Key Concepts

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

Friction
Friction is the force that resists motion between two surfaces in contact. It plays a crucial role when moving objects, such as a crate, across a floor. There are two types of friction relevant here: static and kinetic friction. When an object is at rest, static friction holds it in place against external forces. To overcome this force, the applied force must be greater than the maximum static friction force. Once an object begins moving, kinetic friction takes over, which is usually less than static friction, allowing the object to continue moving with less force.
Understanding these types influences how much force needs to be applied to initiate and maintain movement across a surface.
Inclined Planes
Inclined planes are flat surfaces tilted at an angle. They are ubiquitous in physics problems, altering how forces act on an object. When a force is applied along an inclined plane, not all of the force contributes to moving the object up. Part of it counters gravity, and the rest moves the object. For a crate pulled by a rope inclined above a plane, the force direction must be considered in both horizontal and vertical aspects.
This altered orientation changes how normal force and friction equations are set, modifying what might seem a straightforward force calculation.
Force Components
Force components come into play whenever a force acts at an angle. They break down a complex force into parts that are easier to analyze. For the crate problem, the force applied by the rope can be split into horizontal and vertical components using trigonometry:
  • Horizontal Component: This part works against the friction force to aid motion. It's calculated by multiplying the applied force by the cosine of the angle above the horizontal.
  • Vertical Component: This influences the normal force 鈥� the opposing force exerted by the surface 鈥� calculated via the sine of the angle.
These components are crucial in determining how much force is necessary to overcome friction and begin movement.
Static and Kinetic Friction
Static and kinetic friction are distinct forces acting on objects in different states of motion. Static friction, represented by \(\mu_s\), keeps an object at rest until a threshold force is applied. In the crate example: \[f_s = \mu_s N\] where \(N\) is the adjusted normal force.
Once the object moves, kinetic friction, noted with \(\mu_k\), takes over. Kinetic friction is generally less than static friction, making it easier to keep the object in motion than to start its movement. For moving crates, kinetic friction is captured as: \[f_k = \mu_k N\]. The shift from static to kinetic friction is a pivotal concept in dynamics, crucial for solving motion-related problems on surfaces.

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Most popular questions from this chapter

mechanical engineers must consider how small variations in certain parameters can alter the net force on a passenger. Consider a passenger of mass \(m\) riding around a horizontal circle of radius \(r\) at speed \(v\). What is the variation \(d F\) in the net force magnitude for (a) a variation \(d r\) in the radius with \(v\) held constant, (b) a variation \(d v\) in the speed with \(r\) held constant, and (c) a variation \(d T\) in the period with \(r\) held constant?

ssM A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg}\). Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the net force on the bicycle from the road.

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B\), which was stopped at a red light along a road headed down a hill (Fig. 6-25). You find that the slope of the hill is \(\theta=12.0^{\circ}\), that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{0}=18.0 \mathrm{~m} / \mathrm{s}\). With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) \(0.60\) (dry road surface) and (b) \(0.10\) (road surface covered with wet leaves)?

(co A \(3.5 \mathrm{~kg}\) block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

ssM A car weighing \(10.7 \mathrm{kN}\) and traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) without negative lift attempts to round an unbanked curve with a radius of \(61.0 \mathrm{~m} .(\mathrm{a})\) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is \(0.350\), is the attempt at taking the curve successful?
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