The magnitude of two forces acting on an object are 110 pounds, S60°E, and 120 pounds, N53°E, respectively. Find the magnitude, to the nearest hundredth of a pound, and the direction of the angle, to the nearest tenth of a degree, of the resultant force.

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The magnitude of two forces acting on an object are 110 pounds, S60°E, and 120 pounds, N53°E, respectively.

The forces can be divided into their components in the north-south and east-west axis.

The force of 110 pounds S 60 degrees East has a component of 110*sin 60 towards the East and a component of 110*cos 60 towards the South.

The force 120 pounds N 53 degrees East has a component 120*sin 53 towards the East and 120*cos 53 towards the North.

The sum of the two forces has a component : 120*cos 53 - 110*cos 60 = 17.2178 pounds towards the North and a component 110*sin 60 + 120*sin 53 = 191.099 towards the East.

This gives the magnitude of the combined forces as `sqrt(17.2178^2 + 191.099^2)` = 191.873 pounds. The direction of the force is at an angle of `theta` to the North towards the East. `theta` = `tan^-1(191.099/17.2178)` = 84.85 degrees.

**The resultant force is 191.873 pounds in the direction North 84.85 degrees East.**

If I had 90lbs, S66°E, and 50 lbs, N62°E then

mgnitude = √ ( ( ( ( 90 * ( cos ( - 66 ° ) ) + 50 * ( cos ( 62 ° ) ) ) ^ 2 + ( ( 90 * ( sin ( - 66 ° ) ) ) + ( 50 * ( sin ( 62 ° ) ) ) ) ^ 2 ),

right?

And then that would become

= √ ( ( 36.60629787 + 23.473578139) ^ 2 + ( - 82.2190911878 + 43.301270189221 ) ^ 2 )

= √ ( 60.079876009 ^ 2 + -39.82782099 ^ 2 )

= √ 5195.846826751664

**magnitude = 72.082222 pounds** and

angle = tan ^ -1 ( 60.079876009 / -39.82782099 )

**angle = 54.46°, or N54.46°E.**

Correct? Thanks.

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