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3. A crate is pushed in the positive direction along a rough, horizontal surface with a force F that makes an angle e 45.00 below the horizontal. The magnitude of the force applied is F-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between the crate and the surface is 0.300 45° a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin of each force. (example: Fe (Earth pulling block)) FN 294141 = 435 N FN: Surface on crate Fr: String on crate FG: Earth on crate FK: Surface on crate Ex 200cos45 =141 N FK UKFN 130.5 N 450 FG=mg 30x9.8 = 294 Use 3 significant figures! F Ex=200sin45 =141 N b) Find the acceleration of the crate EFy 0FN = mg + Fsin45° = 294 + 141 = 435 N EFx Fcos450 - Fk 200 cos450 - 0.300*435 = 141 130.5 = 10.5 N EFx ma a = EFx /m = 10.5 /30.0 = 0.350 m/s2

3. A crate is pushed in the positive direction along a rough, horizontal surface with a force F that makes an angle e 45.00 below the horizontal. The magnitude of the force applied is F-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between the crate and the surface is 0.300 45° a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin of each force. (example: Fe (Earth pulling block)) FN 294141 = 435 N FN: Surface on crate Fr: String on crate FG: Earth on crate FK: Surface on crate Ex 200cos45 =141 N FK UKFN 130.5 N 450 FG=mg 30x9.8 = 294 Use 3 significant figures! F Ex=200sin45 =141 N b) Find the acceleration of the crate EFy 0FN = mg + Fsin45° = 294 + 141 = 435 N EFx Fcos450 - Fk 200 cos450 - 0.300*435 = 141 130.5 = 10.5 N EFx ma a = EFx /m = 10.5 /30.0 = 0.350 m/s2

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter6: Applications Of Newton's Laws
Section: Chapter Questions
Problem 48P: Suppose you have a 120-kg wooden crate resting on a wood floor, with coefficient of static friction...
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My question is why did we used sine and cosine 45 degree? Is not is supposed to be 315 degree since 45 degree is in quadrant four? Why can't we use 315 degree?

3. A crate is pushed in the positive direction along a rough, horizontal surface with a force F that makes an angle e 45.00 below the horizontal. The magnitude of the force applied is F-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between the crate and the surface is 0.300 45° a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin of each force. (example: Fe (Earth pulling block)) FN 294141 = 435 N FN: Surface on crate Fr: String on crate FG: Earth on crate FK: Surface on crate Ex 200cos45 =141 N FK UKFN 130.5 N 450 FG=mg 30x9.8 = 294 Use 3 significant figures! F Ex=200sin45 =141 N b) Find the acceleration of the crate EFy 0FN = mg + Fsin45° = 294 + 141 = 435 N EFx Fcos450 - Fk 200 cos450 - 0.300*435 = 141 130.5 = 10.5 N EFx ma a = EFx /m = 10.5 /30.0 = 0.350 m/s2
Transcribed Image Text:3. A crate is pushed in the positive direction along a rough, horizontal surface with a force F that makes an angle e 45.00 below the horizontal. The magnitude of the force applied is F-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between the crate and the surface is 0.300 45° a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin of each force. (example: Fe (Earth pulling block)) FN 294141 = 435 N FN: Surface on crate Fr: String on crate FG: Earth on crate FK: Surface on crate Ex 200cos45 =141 N FK UKFN 130.5 N 450 FG=mg 30x9.8 = 294 Use 3 significant figures! F Ex=200sin45 =141 N b) Find the acceleration of the crate EFy 0FN = mg + Fsin45° = 294 + 141 = 435 N EFx Fcos450 - Fk 200 cos450 - 0.300*435 = 141 130.5 = 10.5 N EFx ma a = EFx /m = 10.5 /30.0 = 0.350 m/s2
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