MHB What is the resultant force and angle for two forces acting on an object?

  • Thread starter Thread starter RobertoPink
  • Start date Start date
  • Tags Tags
    Forces
Click For Summary
SUMMARY

The discussion focuses on calculating the resultant force and angle for two forces acting on an object: a 100-pound force at a 40-degree angle and a 150-pound force along the positive x-axis. Using the method of components, the x-component of the resultant force is calculated as \(R_x = 150 + 100\cos(40)\), and the y-component as \(R_y = 100\sin(40)\). The magnitude of the resultant force is determined using \(|R| = \sqrt{R_x^2 + R_y^2}\), and the angle of the resultant force relative to the positive x-axis is found using \(\theta = \arctan\left(\dfrac{R_y}{R_x}\right)\).

PREREQUISITES
  • Understanding of vector components in physics
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Familiarity with the concept of resultant forces
  • Ability to perform basic algebraic calculations
NEXT STEPS
  • Study vector addition and subtraction techniques
  • Learn about the law of cosines and its applications in physics
  • Explore graphical methods for vector representation
  • Investigate the effects of different angles on resultant forces
USEFUL FOR

Students in physics, engineers working with forces, and anyone interested in understanding vector mechanics and resultant force calculations.

RobertoPink
Messages
2
Reaction score
0
A 100 pound force and a 150 pound force act on an object. If the angle between the force 40 degress, find the magnitude of the resultant force and the angle the resultant force makes with the 150 pound force.
 
Mathematics news on Phys.org
RobertoPink said:
A 100 pound force and a 150 pound force act on an object. If the angle between the force 40 degress, find the magnitude of the resultant force and the angle the resultant force makes with the 150 pound force.

let the 150 lb force be directed along the positive x-axis and the 100 lb force be directed 40 deg relative to the positive x-axis in quadrant I (recommend you make a sketch)

using the method of components in the x and y directions, $R$ represents the resultant vector, $\theta$ is the resultant vector's direction relative to the positive x-axis ...

$R_x = 150 + 100\cos(40)$

$R_y = 0 + 100\sin(40)$

$|R| = \sqrt{R_x^2+R_y^2}$

$\theta = \arctan\left(\dfrac{R_y}{R_x}\right)$
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
6K
Replies
4
Views
6K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
4
Views
2K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K