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ssm A $325-\mathrm{kg}$ boat is sailing $15.0^{\circ}$ north of east at a speed of 2.00 $\mathrm{m} / \mathrm{s}$ . Thirty seconds later, it is sailing $35.0^{\circ}$ north of east at at a speed of 4.00 $\mathrm{m} / \mathrm{s}$ . During this time, three forces act on the boat: a $31.0 \mathrm{N}$ force directed $15.0^{\circ}$ north of east (due to an auxiliary engine), a $23.0-\mathrm{N}$ force directed $15.0^{\circ}$ south of west (resistance due to the water), and $\overrightarrow{\mathbf{F}}_{\mathrm{w}}$ (duc to the wind). Find the magnitude and direction of the force $\overrightarrow{\mathbf{F}}_{\mathrm{W}}$ . Express the direction as an angle with respect to due east.

$F_{W}=19.097 N$ angled $65.9^{\circ}$ North of East

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

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we begin this question by calculating the acceleration off the boat so before a T equals zero its velocity waas 15 degrees north off the east. So let us. They composed this velocity in its acts and why components and I'm choosing X and y components as follows. These will be my why direction in these will be my ex direction. Therefore, the X component off the velocity AT T equals zero. Easy question. True time times the co sign off 15 degrees Because the east direction coincides with the extraction on the white component will be close to two times the sign ar 15 degrees. No at T equals 30. We have the following the axe component off the velocity is eco's too four times the co sign off 35 degrees. The Y component is he goes to four times the sign off 35 degrees. Then we can now calculate acceleration. During these 30 cycles, we have the following the X acceleration. Is it close to the variation in the X component off the velocity divided by the time interval. So we have four times the co sign off 35 degrees minus two times the co sign off 15 degrees divided by 30 seconds. These is approximately zero 0.0 45 meters per second squared. No for the white component. We have the following the variation in the white components off the velocity, divided by the time interval and these egos to four times the sign off 35 degrees, minus two times that sign off 15 degrees. You write it by 30 seconds, which he's approximately 0.0 59 meters per second squared. Now let me organize the board. Okay, now we calculate net force in each direction x and y, and it goes as follows through on the X direction. The Net Force is equal to the following its secrets to the wind force on the X direction, which we don't know yet. Plus 31 times they co sign off 15 degrees, which is the X component off these force miners 23 times the co sign off 15 degrees, which is the ex component off these force. It's negative because it's pointing to the left. This gives us X component off the net force as f w x plus seven 0.7 27 for the white direction we have the following the net force on the white direction. It's equals to the wind force on the wind direction plus 31 times this sign off 15 degrees, minus 23 times this sign ar 15 degrees. This gives us a net force from the right direction off F W white plus 2.71 Now let me organize the board again. Fine. Now that we know both the net forces and acceleration in each direction, we can use Newton's second law to complete. What are the components of doing forced? It goes as follows for the X component we have that the net force on next direction is equal to the mass. Off the mood Times Tex Elevation No ex direction Sure F W x plus 7.7 27 Is he close to 325 kilograms? Times 0.0 45. So the X component off the wind force is equal to 325 times 0.0 45 minus 7.7 27 and this is approximately 6.9 new tones now, repeating the idea for the white component. We get the net force on the white direction is it caused the mass times acceleration in the wind direction. So after w why plus 2.71 is he close to 325 times 0.0 59? Then f W Why is he questioned? 325 times 0.0 59 miners. True 0.7 to 1, which is approximately 17 0.1 Newton. Now let me work in as the board again. Now that we knew both components off the wind force, we can combine them to calculate the magnitude of the wind force using attack. Warren Feur Um, we get a win force that is equal to the square it off F W x squared plus F W y square. Then F W is equal to the square it off 6.9 square plus 17.1 squared, which is approximately 18.4 new terms. So this is the magnitude off the wind force now that is, calculate its direction. So the tangent off the angle that the wind force makes with the X axis Easy clothes, too. The why component off the wind force divided by the X component off the main force, which is close to 17.1, divided by 6.9. And these is approximately true 0.5. Then the angle that the wind force makes with the X axis is equals to the inverse tangent off, too. 0.5. And this gives us an angle off approximately 68. The grease two. The wind force can be drawn as follows. Here is the east direction and then we have a wind force off. 18 went for new toes 0.68 degrees north off the east, So 18.4 noodles force pointing 68 degrees north off the east.

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