/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 A man is dragging a trunk up the... [FREE SOLUTION] | 91Ó°ÊÓ

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A man is dragging a trunk up the loading ramp of a mover's truck. The ramp has a slope angle of \(20.0^{\circ},\) and the man pulls upward with a force \(\vec{F}\) whose direction makes an angle of \(30.0^{\circ}\) with the ramp (Fig. E4.4). (a) How large a force \(\vec{F}\) is necessary for the component \(F_{x}\) parallel to the ramp to be \(90.0 \mathrm{~N} ?\) (b) How large will the component \(F_{y}\) perpendicular to the ramp be then?

Short Answer

Expert verified
The required force \(F\) is \(103.9 N\) and the component of \(F\) perpendicular to the slope or \(F_y\) is \(51.9N\).

Step by step solution

01

Calculate the vector components

We know: \(F_{x} = F*cos(30.0^\circ)\). From this, we can solve for \(F\), which gives: \(F = F_{x} / cos(30.0^\circ)\). Substituting for \(F_{x} = 90.0 N\), we find: \(F = 90.0 N / cos(30.0^\circ)\)
02

Solve for \(\vec{F}\)

To find the size of the force \(\vec{F}\), we substitute the cosine of \(30.0^\circ\) which is \(0.866\). So, \(F = 90.0 N / 0.866 = 103.9 N\).
03

Calculate \(F_{y}\)

Now we can find the perpendicular component \(F_{y}\) using \(F_{y} = F*sin(30.0^\circ)\). Substituting for \(F = 103.9 N\), we get \(F_{y} = 103.9 N * 0.5 = 51.9 N\).

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Key Concepts

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

Vector Components in Physics
Understanding vector components is essential when analyzing forces in physics. A vector represents a quantity that has both magnitude and direction. For example, the force \( \vec{F} \) that the man applies to the trunk in the ramp problem has a specific direction and strength. To simplify analysis, vectors are often broken down into components along the x and y axes, which correspond to the horizontal and vertical directions, respectively.

In our ramp scenario, the force's horizontal component (parallel to the ramp) is denoted as \( F_x \) and is found by multiplying the force's magnitude by the cosine of the angle between the force and the ramp: \( F_x = F \cdot \cos(\theta) \) where \( \theta \) is the angle of interest. In the given problem, we seek the force \( F \) such that its component along the ramp \( F_x \) equals 90.0 N.

Vectors and their components allow us to solve complex problems by focusing on one dimension at a time, making it easier to apply Newton's laws of motion and other principles. By doing so, we can find solutions without having to consider the entire vector all at once.
Trigonometry in Physics
Trigonometry plays a vital role in physics, particularly when dealing with angles and vector components. In our ramp problem, trigonometric functions such as sine and cosine are used to resolve the applied force into components parallel and perpendicular to the ramp's surface.

The cosine function relates the adjacent side of a right-angled triangle to the hypotenuse, which we use to find \( F_x \) as shown in the solution steps. Specifically, \( \cos(30.0^\circ) \) is the ratio of the horizontal component to the total force when the vector makes a 30-degree angle with the horizontal.

Similarly, the sine function, which relates the opposite side to the hypotenuse, helps us determine the vertical component \( F_y \) of the force. Applying the sine function to our example, \( F_y = F \cdot \sin(30.0^\circ) \) gives us the magnitude of the force acting perpendicular to the ramp. Understanding how to use these basic trigonometric functions is crucial for analyzing forces and motion in two dimensions.
Force Parallel and Perpendicular
In physics problems, especially those involving inclined planes like ramps, it's crucial to distinguish between the components of force that are parallel and perpendicular to the surface. When an object is being moved along a ramp, the force parallel to the ramp, \( F_x \), influences the object's motion up or down the incline.

In the case of the trunk being pulled up the ramp by the man in our exercise, we use trigonometry to determine the necessary parallel force. This force works against gravity and any friction present on the ramp’s surface. In contrast, the perpendicular force component, \( F_y \), is related to the normal force, which does not contribute to moving the object along the ramp but affects the pressure between the object and ramp surface.

The greater the perpendicular force, the more significant the normal force and potential frictional force could be. Since friction often depends on the normal force, \( F_y \) can alter how hard it is to slide the trunk up or down the ramp. Grasping the interplay of these force components is key to understanding real-world applications, such as the work done by the man or the energy needed to overcome friction.

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Most popular questions from this chapter

Boxes \(A\) and \(B\) are in contact on a horizontal, frictionless surface (Fig. E4.23). Box A has mass \(20.0 \mathrm{~kg}\) and box \(B\) has mass \(5.0 \mathrm{~kg} .\) A horizontal force of \(250 \mathrm{~N}\) is exerted on box \(A\). What is the magnitude of the force that box \(A\) exerts on box \(B ?\)

A block of mass \(2.00 \mathrm{~kg}\) is initially at rest at \(x=0\) on a slippery horizontal surface for which there is no friction. Starting at time \(t=0,\) a horizontal force \(F_{x}(t)=\beta-\alpha t\) is applied to the block, where \(\alpha=6.00 \mathrm{~N} / \mathrm{s}\) anwd \(\beta=4.00 \mathrm{~N}\). (a) What is the largest positive value of \(x\) reached by the block? How long does it take the block to reach this point, starting from \(t=0,\) and what is the magnitude of the force when the block is at this value of \(x ?\) (b) How long from \(t=0\) does it take the block to return to \(x=0,\) and what is its speed at this point?

Compared with the force her neck exerts on her head during the landing, the force her head exerts on her neck is (a) the same; (b) greater; (c) smaller; (d) greater during the first half of the landing and smaller during the second half of the landing.

A batter swings at a baseball (mass \(0.145 \mathrm{~kg}\) ) that is moving horizontally toward him at a speed of \(40.0 \mathrm{~m} / \mathrm{s} .\) He hits a line drive with the ball moving away from him horizontally at \(50.0 \mathrm{~m} / \mathrm{s}\) just after it leaves the bat. If the bat and ball are in contact for \(8.00 \mathrm{~ms}\), what is the average force that the bat applies to the ball?

A hockey puck with mass \(0.160 \mathrm{~kg}\) is at rest at the origin \((x=0)\) on the horizontal, frictionless surface of the rink. At time \(t=0\) a player applies a force of \(0.250 \mathrm{~N}\) to the puck, parallel to the \(x\) -axis; she continues to apply this force until \(t=2.00 \mathrm{~s}\). (a) What are the position and speed of the puck at \(t=2.00 \mathrm{~s} ?\) (b) If the same force is again applied at \(t=5.00 \mathrm{~s},\) what are the position and speed of the puck at \(t=7.00 \mathrm{~s} ?\)

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