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qz2sol_3081sp10

qz2sol_3081sp10 - is along the y axis thus it has y...

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TA: Tomoyuki Nakayama Monday, January 25 th , 2010 PHY 2048: Physic 1, Discussion Section 3081 Quiz 2 (Homework Set #3) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ There are two forces on the 4.00-kg box in the overhead view of the figure below, but only one is shown, where F 1 = 50.0 N, a = 8.00 m/s and θ = 30.0º. a) Express a and F 1 in unit-vector notation. The angle between a and the +x axis is 30º + 180º = 210º Thus a is expressed as a = acos210º i + asin210º j = -6.92 m/s 2 i - 4.00 m/s 2 j F 1
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Unformatted text preview: is along the y axis, thus it has y component only. F 1 = 50.0 N j b) Find the second force in unit-vector notation. We apply the Newton’s 2 nd law to the block and solve it for F 2 . We get m a = F 1 + F 2 ⇒ F 2 = m a-F 1 = 4 × (-6.92 i - 4.00 j ) -50.0 j = -27.7 N i-66.0 N j c) What are the magnitude and direction (relative to the + x axis) of the second force? Using Pythagorean Theorem, the magnitude is F 2 = √(F 1x 2 + F 1y 2 ) = √((-27.7) 2 + (-66) 2 ) = 71.6 N The angle is given by the inverse tangent function. Since F 2 is in the 3 rd quadrant, we add 180º to the answer your calculator displays. θ 2 = tan-1 (F 1y /F 1x ) +180º = 247º...
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