Answer :
The direction of the resultant force is 74.2° counterclockwise from the positive x-axis and the magnitude of the resultant force is 243 pounds.
Let F1, F2, and F3 be the three forces having magnitudes of 64, 114, and 140 pounds, respectively. The angles between the forces and the positive x-axis are 30°, 45°, and 120°. We will apply the head-to-tail vector addition method to find the direction and magnitude of the resultant force. The steps are as follows:
Let's begin the solution by drawing the vectors to scale in a coordinate system as shown below:
Step 1: Drawing the vectors to scale.We can now measure the components of each vector.
The x and y components of each vector are:
F1 = 64 pounds at 30°x-component:
64 cos 30° = 55.43 pounds
y-component: 64 sin 30° = 32 pounds
F2 = 114 pounds at 45°x-component:
114 cos 45° = 80.5 pounds
y-component: 114 sin 45° = 80.5 pounds
F3 = 140 pounds at 120°x-component:
140 cos 120° = -70 pounds
y-component: 140 sin 120° = 121.11 pounds
Adding the vectors head to tail.We can now add the vectors head to tail, starting with F1 and F2, and then adding F3 to the sum of F1 and F2. The vectors are drawn to scale below: The resultant vector is drawn from the origin to the head of the last vector in the chain.
Drawing the resultant vector.The direction of the resultant vector is measured counterclockwise from the positive x-axis. We can measure this angle by using the tangent function as follows:
θ = tan⁻¹(ΣFy/ΣFx)
θ = tan⁻¹[(32 + 80.5 + 121.11)/(55.43 + 80.5 - 70)]
θ = tan⁻¹(233.61/66.93)
θ = 74.2°
The magnitude of the resultant vector is given by:
R = √(ΣFx² + ΣFy²)
R = √[(55.43 + 80.5 - 70)² + (32 + 80.5 + 121.11)²]
R = √(222.57² + 233.61²)
R = 243 pounds
Therefore, the direction of the resultant force is 74.2° counterclockwise from the positive x-axis and the magnitude of the resultant force is 243 pounds.
Thus, the direction of the resultant force is 74.2° counterclockwise from the positive x-axis and the magnitude of the resultant force is 243 pounds.
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