Question

two forces act on a 55 - kg object. one force has a magnitude 65 n directed 57° clockwise from the positive x - axis, and the other has a magnitude 35 n at 32° clockwise from the positive y - axis. what is the magnitude of this objects acceleration?
a) 1.9 m/s2
b) 1.5 m/s2
c) 1.3 m/s2
d) 1.7 m/s2
e) 1.1 m/s2

Explanation:

Step1: Calculate \(x\) - components of forces

For the first force \(F_1 = 65N\) at \(\theta_1=57^{\circ}\) clockwise from \(+x\) - axis. The \(x\) - component \(F_{1x}=F_1\cos\theta_1\), and for the second force \(F_2 = 35N\) at \(\theta_2 = 90^{\circ}+ 32^{\circ}=122^{\circ}\) from \(+x\) - axis. The \(x\) - component \(F_{2x}=F_2\cos\theta_2\).

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The total \(x\) - component \(F_x=F_{1x}+F_{2x}\approx35.425-18.5465 = 16.8785N\)

Step2: Calculate \(y\) - components of forces

The \(y\) - component of \(F_1\): \(F_{1y}=-F_1\sin\theta_1\) (clockwise from \(+x\) - axis). The \(y\) - component of \(F_2\): \(F_{2y}=-F_2\sin\theta_2\) (clockwise from \(+y\) - axis means clockwise from \(+x\) - axis by \(90 + 32\) degrees).

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The total \(y\) - component \(F_y=F_{1y}+F_{2y}\approx-54.535-29.68=-84.215N\)

Step3: Calculate the magnitude of the resultant force

Using the Pythagorean theorem \(F=\sqrt{F_x^{2}+F_y^{2}}\)

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Step4: Calculate the acceleration

Using Newton's second law \(F = ma\), where \(m = 55kg\) and \(F\approx85.9N\). Then \(a=\frac{F}{m}\)

$$a=\frac{85.9}{55}\approx1.56m/s^{2}\approx1.5m/s^{2}$$

Answer:

B. \(1.5m/s^{2}\)