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Chapter 3: Problem 43

Compute algebraically the resultant of the following coplanar forces: \(100 \mathrm{~N}\) at \(30^{\circ}, 141.4 \mathrm{~N}\) at \(45^{\circ}\), and \(100 \mathrm{~N}\) at \(240^{\circ}\). Check your result graphically.

Short Answer

Expert verified
Resultant force is 150.6 N at 25掳 above the x-axis.

Step by step solution

01

Resolve Forces into Components

We start by resolving each force into its x and y components. For a force \( F \) at an angle \( \theta \):- The x-component is \( F_x = F \cos \theta \).- The y-component is \( F_y = F \sin \theta \).For the 100 N force at \(30^{\circ}\):\[F_{1x} = 100 \cos 30^{\circ} = 86.6 \text{ N}, \F_{1y} = 100 \sin 30^{\circ} = 50 \text{ N}.\]For the 141.4 N force at \(45^{\circ}\):\[F_{2x} = 141.4 \cos 45^{\circ} = 100 \text{ N}, \F_{2y} = 141.4 \sin 45^{\circ} = 100 \text{ N}.\]For the 100 N force at \(240^{\circ}\):\[F_{3x} = 100 \cos 240^{\circ} = -50 \text{ N}, \F_{3y} = 100 \sin 240^{\circ} = -86.6 \text{ N}.\]
02

Sum the Components

Sum all x-components and y-components of the forces to find the resultant components.The resultant x-component (\( R_x \)) is:\[R_x = F_{1x} + F_{2x} + F_{3x} = 86.6 + 100 - 50 = 136.6 \text{ N}\]The resultant y-component (\( R_y \)) is:\[R_y = F_{1y} + F_{2y} + F_{3y} = 50 + 100 - 86.6 = 63.4 \text{ N}\]
03

Calculate the Resultant Force

Use the Pythagorean theorem to find the magnitude of the resultant force:\[R = \sqrt{R_x^2 + R_y^2} = \sqrt{136.6^2 + 63.4^2} \]\[R = \sqrt{18665.56 + 4020.36} = \sqrt{22686.12} = 150.6 \text{ N}\]
04

Determine the Direction of the Resultant

The direction \( \theta \) of the resultant force can be found using:\[\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{63.4}{136.6}\right)\]\[\theta = \tan^{-1}(0.464) \approx 25^{\circ}\]Thus, the direction of the resultant force is \(25^{\circ}\) above the positive x-axis.

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Key Concepts

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

Vector Components
When dealing with forces in physics, a convenient method is to break them down into perpendicular components. This process turns any force that acts at an angle into a pair of one horizontal (x-axis) and one vertical (y-axis) component. These are known as vector components. By doing this, it allows us to simplify complex problems.To resolve a force vector into its components, you will use basic trigonometry:
  • The x-component of a force, represented as \( F_x \), is calculated by multiplying the magnitude of the force \( F \) by the cosine of its angle \( \theta \): \( F_x = F \cos \theta \).
  • The y-component, \( F_y \), is found by multiplying the force magnitude by the sine of the angle: \( F_y = F \sin \theta \).
Using these equations, any force can be separated into parts that align with the standard Cartesian coordinate system, making the mathematics behind force analysis much more manageable.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry that is often used in physics to determine the resultant magnitude of two perpendicular forces. When you have resolved vectors into components, the next step is to find the resultant vector. The theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In mathematical terms:\[ c^2 = a^2 + b^2 \]In physics, if \( R_x \) and \( R_y \) are the resultant x and y components respectively, then the magnitude of the resultant force \( R \) is determined by:\[ R = \sqrt{R_x^2 + R_y^2} \]This equation lets you calculate the single force that results from the combination of all forces acting perpendicularly, providing both simplification and clarity in analyzing vector quantities.
Trigonometry in Physics
Trigonometry is a crucial mathematical discipline used extensively in physics to determine unknown quantities in systems involving angles. When calculating forces, trigonometry bridges the gap between forces acting along different directions and enables us to predict outcomes accurately.When you resolve forces into vector components as covered earlier, you'll employ trigonometric functions:
  • Cosine helps identify the adjacent side of a force vector, which in physics corresponds to the x-component.
  • Sine, on the other hand, identifies the opposite side, or y-component, relevant when working with vectors.
  • Tangent is used for angles, as in computing the angle of a resultant force: \( \tan \theta = \frac{opposite}{adjacent} = \frac{R_y}{R_x} \).
Mastering trigonometry provides a toolkit for predicting behavior in physical systems where angles and distances play a role. This makes it indispensable for calculating trajectories, forces, and other essential physics properties.
Angle of Resultant Force
Determining the angle of the resultant force is pivotal as it helps you understand the direction in which the resultant force is acting. Once the x and y components are determined and summed, the direction or angle of this resultant force can reveal the net effect of all individual forces. To find this angle, we use the tangent function, which relates the ratio of the opposite side to the adjacent side in a right triangle formed by the force components. Thus, you can find the angle \( \theta \) using:\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \]This inverse tangent function calculates the angle formed between the resultant vector and the positive x-axis. Understanding this angle gives you insight into the directional impact of combined forces, ensuring you correctly specify how these forces influence real-world scenarios, like motion direction or structural stability.

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Most popular questions from this chapter

A \(400-g\) block with an initial speed of \(80 \mathrm{~cm} / \mathrm{s}\) slides along a horizontal tabletop against a friction force of \(0.70 \mathrm{~N}\). (a) How far will it slide before stopping? \((b)\) What is the coefficient of friction between the block and the tabletop? (a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, \(-0.70 \mathrm{~N}\). Therefore, $$ \sum F=m a \quad \text { becomes } \quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a) $$ from which \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). (Notice that \(m\) is always in kilograms.) To find the distance the block slides, make use of \(v_{i x}=0.80 \mathrm{~m} / \mathrm{s}, v_{f x}=0\), and \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). Then \(v_{f x}^{2}-v_{i x}^{2}=2 a x\) yields $$ x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m} $$ (b) Because the vertical forces on the block must cancel, the upward push of the table \(F_{N}\) must equal the weight \(m g\) of the block. Then $$ \mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18 $$

A \(600-\mathrm{kg}\) car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. (a) First find the car's acceleration from a constant- \(a\) equation. It is known that \(v_{i x}=30 \mathrm{~m} / \mathrm{s}, v_{f x}=0\), and \(x=70 \mathrm{~m}\). Use \(v_{f x}^{2}=v_{i x}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in ( \(a\) ) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(\mu_{s}=F_{\mathrm{f}} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, $$ F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} $$ so that $$ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

A 200-N wagon is to be pulled up a \(30^{\circ}\) incline at constant speed. How large a force parallel to the incline is needed if friction effects are negligible? The situation is shown in Fig. \(3-16(a)\). Because the wagon moves at a constant speed along a straight line, its velocity vector is constant. Therefore, the wagon is in translational equilibrium, and the first condition for equilibrium applies to it. We isolate the wagon as the object. Three non-negligible forces act on it: (1) the pull of gravity \(F_{W}\) (its weight), directed straight down; (2) the applied force \(F\) exerted on the wagon parallel to the incline to pull it up the incline; (3) the push \(F_{N}\) of the incline that supports the wagon. These three forces are shown in the free-body diagram in Fig. \(3-16\). For situations involving inclines, it is convenient to take the \(x\) -axis parallel to the incline and the \(y\) -axis perpendicular to it. After taking components along these axes, we can write the first condition for equilibrium: $$ \begin{array}{lll} +\sum F_{x}=0 & \text { becomes } & F-0.50 F_{W}=0 \\ +\sum F_{y}=0 & \text { becomes } & F_{N}-0.866 F_{W}=0 \end{array} $$ Solving the first equation and recalling that \(F_{W}=200 \mathrm{~N}\), we find that \(F=0.50 F_{W}\). The required pulling force to two significant figures is \(0.10 \mathrm{kN}\).

A cord passing over a light frictionless pulley has a \(7.0\) -kg mass hanging from one end and a \(9.0-\mathrm{kg}\) mass hanging from the other, as seen in Fig. 3-20. (This arrangement is called Atwood's machine.) Find the acceleration of the masses and the tension in the cord. Because the pulley is easily turned, the tension in the cord will be the same on each side. The forces acting on each of the two masses are drawn in Fig. 3-20. Recall that the weight of an object is \(m g\). It is convenient in situations involving objects connected by cords to take the overall direction of motion of the system as the positive direction. That's often indicated by the direction of motion of the pulley when the system is let free to move. In the present case, the pulley would turn clockwise, and so we take \(u p\) positive for the \(7.0\) -kg mass, and down positive for the \(9.0\) -kg mass. (If we do this, the acceleration will be positive for each mass. Because the cord doesn't stretch, the accelerations are numerically equal.) Writing \(\sum F_{y}=m a_{y}\) for each mass in turn, $$ +\uparrow \sum F_{y A}=F_{T}-(7.0)(9.81) \mathrm{N}=(7.0 \mathrm{~kg})(a) \text { and }+\downarrow \sum F_{y B}=(9.0)(9.81) \mathrm{N}-F_{T}=(9.0 \mathrm{~kg})(a) $$ Add these two equations and the unknown \(F_{T}\) drops out, giving $$ (9.0-7.0)(9.81) \mathrm{N}=(16 \mathrm{~kg})(a) $$ for which \(a=1.23 \mathrm{~m} / \mathrm{s}^{2}\) or \(1.2 \mathrm{~m} / \mathrm{s}^{2}\). Now substitute \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) for \(a\) in either equation and obtain \(F_{T}=77 \mathrm{~N}\).

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?
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