corrected_problems

corrected_problems - r F1 y r F3 r F2 30o 60o x Define x...

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Unformatted text preview: r F1 y r F3 r F2 30o 60o x Define x and y axes. Then, the net force in the x direction is Fx = ∑i Fxi and the net force in the y Question 1A: An object on a horizontal table is acted on r r r by three forces, F1 = 4 N, F2 = 3N and F3 = 5N, as shown direction is Fy = ∑i Fyi . ∴ Fx = F1 cos 30o + F2 cos 60o − F3 = 3.46 + 1.50 − 5.00 = −0.04 N, below. What is the net force acting on the object? r F1 r F3 and Fy = F1 sin 30o + F2 sin 60o 30o r F2 60o = 2.00 − 2.60 = −0.60 N. Fx ˆ i θ r FR Thus, the net force is r F = ( −0.04ˆ − 0.60ˆ ) N, i j Fy ˆ j R with magnitude r FR = (Fx )2 + (Fy )2 = ( −0.04)2 + ( −0.60)2 = 0.601N . The angle θ = tan −1 Fy = tan −1 (15) = 86.2o . Fx 1 r F1 y r F3 r F2 Question 1B: An object on a horizontal table is acted on r r r by three forces, F1 = 4 N, F2 = 3N and F3 = 5N, as shown below. If the mass of the object is 2.0kg, what is it’s acceleration? r F1 r F3 30o r F2 60o 30o 60o x From before (question 1A) we found the resultant force: r FR = ( −0.04ˆ − 0.60ˆ ) N. i j r r Using Newton’s 2nd Law FR = ma . r i j r FR ( −0.04ˆ − 0.60ˆ ) ∴a = = 2 m = ( −0.02ˆ − 0.30ˆ ) m/s2 , i j and so the magnitude of the acceleration is r a = ( −0.02)2 + ( −0.30)2 = 0.30 m/s2. r r Note that a and FR are parallel. Hence, if we know the initial position (or velocity) of the object, we can determine its position (or velocity) at some other time by using the equations of motion from chapters 2 and 3. 2 ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.

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