Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 623 north of west. What is the magnitude of the bodys acceleration?
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Question
A 2.00 kg object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{N}) \hat{\mathrm{i}}+(16.0 \mathrm{N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=-(12.0 \mathrm{N}) \hat{i}+(8.00 \mathrm{N}) \hat{j}\), find the third force.
Solution
The first step in solving 5 problem number 8 trying to solve the problem we have to refer to the textbook question: A 2.00 kg object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{N}) \hat{\mathrm{i}}+(16.0 \mathrm{N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=-(12.0 \mathrm{N}) \hat{i}+(8.00 \mathrm{N}) \hat{j}\), find the third force.
From the textbook chapter you will find a few key concepts needed to solve this.
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