The following four forces act on a 4.00 kg object: ₁ = 300 N east
₂ = 700 N north
₃ = 500 N west
₄ = 600 N south
What is the acceleration of the object?

Question

The following four forces act on a 4.00 kg object: ₁ = 300 N east ₂ = 700 N north ₃ = 500 N west ₄ = 600 N south What is the acceleration of the object?

Quizplus · Accepted Answer

To solve the problem, we need to find the net force acting on the object and then use Newton's second law (F=ma) to find the acceleration. To find the net force, we need to add the vector components of the four forces, which form a closed triangle as shown below:

[asy] draw((0,0)--(2.72,-2.08)); draw((2.72,-2.08)--(1.72,-2.08)); draw((1.72,-2.08)--(0,0)); draw((0,0)--(0,-2.08),dashed); draw((0,0)--(3.5,0),Arrow); draw((0,0)--(0,2),Arrow); label("$700$",(1.22,-2.08),S); label("$500$",(-0.86,-1.04),NE); label("$300$",(1.75,0.65),NE); label("$600$",(-1.75,-0.6),W); label("$	heta$",(0.85,-0.15),NE); [/asy]

Using the sine rule, we can find the magnitude of the net force:

$\frac{\sin	heta}{700}=\frac{\sin(180^\circ-90^\circ-	heta)}{600}=\frac{\sin 90^\circ}{300}$

$\sin	heta=\frac{700}{1200}$

$	heta\approx44.5^\circ$

The direction of the net force is 180°-63.4°=116.6° (measured from the west direction). To find the magnitude of the net force, we can use the cosine rule:

$F_{net}^2=700^2+600^2-2\cdot700\cdot600\cdot\cos(116.6^\circ)$

$F_{net}=224	ext{ N}$

Finally, we can use Newton's second law to find the acceleration:

$F_{net}=ma$

$a=\frac{F_{net}}{m}=\frac{224}{4}=56	ext{ m/s}^2$

The acceleration is in the direction of the net force, which is 180°-63.4°=116.6° west of north. To express the direction in standard compass notation, we subtract this angle from 90°:

$90^\circ-116.6^\circ=26.6^\circ$

Therefore, the acceleration is 224 N in a direction 26.6° north of west, so the correct answer is (D).

The Following Four Forces Act on a 4