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Three forces act on a point object. Force 1 has magnitude 30 N at 45 degrees. Force 2 has magnitude 40 N at 0 degrees. Force 3 has magnitude 25 N at 270 degrees. Determine the Cartesian components, ma

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2 parts
b

(53.379 N)i^−(13.287 N)j^(53.379 \, \text{N})\hat{i} - (13.287 \, \text{N})\hat{j}

c

55.008 N55.008 \, \text{N}

Step-by-step solution

3 steps

Here are the next parts of Question 1:

Part b· 2 steps

Determine the Cartesian components of the resultant R of those three forces.

  1. Sum the x-components of all forces.

    Using the values from the previous calculation:

    Rx=F1x+F2x+F3xR_x = F_{1x} + F_{2x} + F_{3x} Rx=21.213 N+38.636 N−6.470 NR_x = 21.213 \, \text{N} + 38.636 \, \text{N} - 6.470 \, \text{N} Rx=53.379 NR_x = 53.379 \, \text{N}

  2. Sum the y-components of all forces.

    Ry=F1y+F2y+F3yR_y = F_{1y} + F_{2y} + F_{3y} Ry=21.213 N−10.352 N−24.148 NR_y = 21.213 \, \text{N} - 10.352 \, \text{N} - 24.148 \, \text{N} Ry=−13.287 NR_y = -13.287 \, \text{N}

    The Cartesian components of the resultant force R⃗\vec{R} are:

    R⃗=(53.379 N)i^−(13.287 N)j^\vec{R} = \boxed{(53.379 \, \text{N})\hat{i} - (13.287 \, \text{N})\hat{j}}

Part c

55.008 N55.008 \, \text{N}

Deduce the magnitude of the resultant R.

  1. Calculate the magnitude of the resultant force.

    The magnitude of the resultant vector is found using the Pythagorean theorem:

    ∣R⃗∣=Rx2+Ry2|\vec{R}| = \sqrt{R_x^2 + R_y^2} ∣R⃗∣=(53.379 N)2+(−13.287 N)2|\vec{R}| = \sqrt{(53.379 \, \text{N})^2 + (-13.287 \, \text{N})^2} ∣R⃗∣=2849.308 N2+176.544 N2|\vec{R}| = \sqrt{2849.308 \, \text{N}^2 + 176.544 \, \text{N}^2} ∣R⃗∣=3025.852 N2|\vec{R}| = \sqrt{3025.852 \, \text{N}^2} ∣R⃗∣≈55.008 N|\vec{R}| \approx \boxed{55.008 \, \text{N}}
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