Question

Page < 37 > of 530 - zoom + Two people pull on a boat with the forces shown below. If the boat is 350 kg and starts from rest, what is the velocity of the boat 5.00 s after the people start pulling it? (Ignore any forces due to the water.) Determine also, the angle the resultant force makes with the x-axis FA = 40.0 N 09 15.00 L FB = 30.0 N Two force vectors act on a boat

Answer 1

FA = 40N at 45° above r - aris

FA= Fari + FAyj = FAcosti + F Asinoj

${\color{Red} F_{A}=28.2843i+28.2843j}$

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FB = 30N at 37% below r - acis

360-37=323

$F_{B}=30N\ at\ 323^{\circ}\ from\ x-axis$

$F_{B}=F_{Bx}i+F_{By}j=F_{B}cos\theta i+F_{B}sin\theta j$

${\color{Red} F_{B}=23.9591i-18.0545j}$

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Resultant force, $F_{R}=F_{A}+F_{B}$

$F_{R}=(28.2843i+28.2843j)+(23.9591i-18.0545j)$

$F_{R}=28.2843i+28.2843j+23.9591i-18.0545j$

$F_{R}=(28.2843+23.9591)i+(28.2843-18.0545)j$

${\color{Red} F_{R}=52.2434i+10.2298j}$

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Magnitude, $F_{R}=\sqrt{(52.2434)^{2}+(10.2298)^{2}}$

${\color{Red} F_{R}=53.2355N}$

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Consider the boat

Net Force = Mass*Acceleration

$F_{R}=ma$

$\frac{F_{R}}{m}=a$

$\frac{53.2355N}{350kg}=a$

${\color{Red} a=0.1521m/s^{2}}$

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Use the formula $v=u+at$ to find the final velocity

$v=0m/s+0.1521m/s^{2}*5s$

$v=0+0.1521*5$

ANSWER : ${\color{Red} v=0.7605m/s}$

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Direction of resultant force

$\theta =tan^{-1}(\frac{F_{Ry}}{F_{Rx}})$

$\theta =tan^{-1}(\frac{10.2298}{52.2434})$

ANSWER : ${\color{Red} \theta =11.08^{\circ}\ above\ the\ x\ axis}$

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