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The only two forces acting on a body have magnitudes of \(20 \mathrm{~N}\) and \(35 \mathrm{~N}\) and directions that differ by \(80^{\circ}\). The resulting acceleration has a magnitude of \(20 \mathrm{~m} / \mathrm{s}^{2}\). What is the mass of the body?

Short Answer

Expert verified
The mass of the body is approximately 2.16 kg.

Step by step solution

01

Understand the Problem

We are given two forces acting on a body, with magnitudes of 20 N and 35 N. The angle between these forces is 80°. The resulting acceleration of the body is 20 m/s². We need to find the mass of the body.
02

Use the Formula for Net Force

First, find the resultant force using the formula for two forces acting at an angle:\[ F_{net} = \sqrt{F_1^2 + F_2^2 + 2 \cdot F_1 \cdot F_2 \cdot \cos \theta} \]Here, \( F_1 = 20 \text{ N} \), \( F_2 = 35 \text{ N} \), and \( \theta = 80^\circ \).
03

Calculate the Cosine of the Angle

Calculate the cosine of 80°:\[ \cos 80^\circ \approx 0.1736 \]
04

Substitute Values and Calculate

Substitute the known values into the net force formula:\[ F_{net} = \sqrt{(20)^2 + (35)^2 + 2 \cdot 20 \cdot 35 \cdot 0.1736} \]Calculate the terms inside the square root:\[ (20)^2 = 400 \]\[ (35)^2 = 1225 \]\[ 2 \cdot 20 \cdot 35 \cdot 0.1736 \approx 242.08 \]Now sum these:\[ 400 + 1225 + 242.08 = 1867.08 \]Finally, calculate the square root:\[ F_{net} \approx \sqrt{1867.08} \approx 43.23 \text{ N} \]
05

Use Newton's Second Law

Apply Newton's Second Law to relate force, mass, and acceleration:\[ F_{net} = m \cdot a \]Solving for mass \( m \):\[ m = \frac{F_{net}}{a} \]Substitute \( F_{net} = 43.23 \text{ N} \) and \( a = 20 \text{ m/s}^2 \):\[ m = \frac{43.23}{20} \]
06

Calculate the Mass

Perform the division to find the mass:\[ m \approx 2.1615 \text{ kg} \]

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Key Concepts

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

Net Force Calculation
To understand how to find net force when two forces act at an angle, it's essential to start with Newton's Second Law and the role of resultant force in motion. When forces are applied to an object, they combine to form a net force, which dictates the motion through acceleration. In our problem, you have two known forces: 20 N and 35 N, with an 80° angle between them.
To calculate the net force, the formula to use is:
\[ F_{net} = \sqrt{F_1^2 + F_2^2 + 2 \cdot F_1 \cdot F_2 \cdot \cos \theta} \]
This accounts for both the magnitudes of the forces and their direction. This is particularly crucial because the forces are not aligned straight with each other.
  • Calculate the force components based on trigonometry.
  • Add the vectors using their components.
  • Use the angle to combine them correctly.
Only through this approach can you find the resultant magnitude that truly impacts the object's acceleration.
Vector Addition of Forces
Vector addition is the backbone of solving problems involving multiple forces. When two or more forces act at an angle, they are not simply added together algebraically, because they influence different dimensions. Instead, you calculate based on directionality using trigonometry.
Each force can be split into components:
  • Horizontal component (using cosine of the angle).
  • Vertical component (using sine of the angle).
For our example, you calculate:
\[ F_x = F_1 \cdot \cos \theta + F_2 \cdot \cos \theta \]
\[ F_y = F_1 \cdot \sin \theta + F_2 \cdot \sin \theta \]
We then combine these components back into a single resultant force using the Pythagorean theorem. This proper method of vector addition allows us to work with forces accurately, especially where angles are involved, and leads to finding the true net force direction and magnitude.
Finding Mass from Force and Acceleration
Once you have the net force calculated, you can find the mass using Newton’s Second Law, which relates mass, acceleration, and force as follows:
\[ F_{net} = m \cdot a \]
Rearranging to find mass, you get:
\[ m = \frac{F_{net}}{a} \]
This step is straightforward once you have the net force and acceleration values. In solving real-world problems:
  • Start with known force and acceleration.
  • Use algebraic manipulation to isolate mass.
  • Input known values to solve for the unknown mass.
In our exercise, the net force calculated was 43.23 N, and with a given acceleration of 20 m/s², you simply divide these to find the mass, yielding approximately 2.1615 kg. This logical sequence from determining forces to calculating mass is pivotal in physics education, making it crucial to practice repeatedly for strong conceptual understanding.

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Most popular questions from this chapter

A \(0.150 \mathrm{~kg}\) particle moves along an \(x\) axis according to \(x(t)=-13.00+2.00 t+4.00 t^{2}-3.00 t^{3}\), with \(x\) in meters and \(t\) in seconds. In unit-vector notation, what is the net force acting on the particle at \(t=3.40 \mathrm{~s}\) ?

An elevator cab and its load have a combined mass of \(1600 \mathrm{~kg}\). Find the tension in the supporting cable when the cab, originally moving downward at \(12 \mathrm{~m} / \mathrm{s}\), is brought to rest with constant acceleration in a distance of \(42 \mathrm{~m}\).

A \(10 \mathrm{~kg}\) monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a \(15 \mathrm{~kg}\) package on the ground (Fig. 5-54). (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) magnitude and (c) direction of the monkey's acceleration and (d) the tension in the rope?

For sport, a \(12 \mathrm{~kg}\) armadillo runs onto a large pond of level, frictionless ice. The armadillo's initial velocity is \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of \(17 \mathrm{~N}\) in the positive direction of the \(y\) axis. In unitvector notation, what are the animal's (a) velocity and (b) position vector when it has slid for \(3.0\) s?

Using a rope that will snap if the tension in it exceeds \(387 \mathrm{~N}\), you need to lower a bundle of old roofing material weighing \(449 \mathrm{~N}\) from a point \(6.1 \mathrm{~m}\) above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

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