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

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To find the resultant force and angle for a 100-pound and a 150-pound force acting at a 40-degree angle, the components of the forces must be calculated. The x-component of the resultant force is derived from the equation R_x = 150 + 100*cos(40), while the y-component is R_y = 100*sin(40). The magnitude of the resultant force is calculated using |R| = √(R_x² + R_y²). The angle of the resultant force relative to the positive x-axis is determined using θ = arctan(R_y/R_x). This method effectively combines the forces to yield the resultant force and its direction.
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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.
 
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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 ...

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

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

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

$\theta = \arctan\left(\dfrac{R_y}{R_x}\right)$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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