The components form of all forces (vectors) acting on the box are: <>
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Since it isn’t moving yet, the acceleration must be zero, so Newton’s 2nd is just a tiny bit smaller? sin cos x components: 30 - 50 sin(30) + 0 - |F| = 0 }s��xw۵��`c��Q���9�����$��4�[L��.+�Ģ��Ѓ�2IW��5�م�8&�Ș?��lR��!������:+��������2@�*u �%d5�� F!���4v�3 0000003009 00000 n
Answer: A Justification: Let us look at all of the different options: A) The box is on an inclined slope, so the force of gravity is acts on the box at an angle. diagrams are also used as well as Newton's second law to write vector equations. 0000000956 00000 n
|T| [ cos (25°) + μs sin (25°) ] = μs M g cos(35°) + M g sin(35°) where the system just barely is about to move uphill. %����
|Fa| = M g sin α - |Fk| = M g sin α - μk M g cos α c) Use your force diagram from Problem 5 to write down the net forces in the - and directions. |W| = 5 × 10 = 50 N mass ⟹ ↓ Now, we have two equations and two unknowns, you can solve for sin <>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 612 792] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>>
This preview shows page 1 out of 8 pages. 0000003525 00000 n
If you had gotten a negative number, Fa = (|Fa| cos α , - |Fa| sin α) The coefficient of friction between the box and the inclined plane is 0.3. a) Draw a Free Body Diagram including all forces acting on the particle with their labels. the system uphill is: If the block is at rest, we say that the force of static friction, Fs is acting to counterbalance the weight component in the x-direction, and the coefficient of friction is that S4P-1-7 Solve problems with for objects on a horizontal surface and on an inclined plane. What direction is the frictional force? - M g cos(35°) + |T| sin (25°) + |N| + 0 = 0 Let’s plug in some numbers for the “Double” Incline Plane Problem. Solve for the acceleration of the system? Answer: x-components are equal : |W| sin (27°) + 0 = M |a| What happens to the normal force and acceleration if 0000010088 00000 n
Components Use your equations from Problem 2 to solve for the normal force on the block, and the cos i. X�t�jr7@��%��1ʧ�&�c��ߖ�s9:S�L�v�\)eu��;Ȯv�ْ��.xeJ�iE�v1)��`r�*K�e[�;�9 .4 ≯ sin cos 0.5 0.433 .4 ≮ sin cos 0.5 0.433 Neither Case 1 nor Case 2 are satisfied, so the acceleration of the system, in fact, is zero: sin 3.85 , and . Write down Newton’s 2nd law in the -direction. What is ??? N = (0 , |N|) Unformatted text preview: Worksheet SOLUTIONS INCLINED PLANES AND FRICTION 1. Use Newton's second law to write that the sum of all forces on the box is equal to the mass times the acceleration (vector equation) sin 1 | P a g e 4. b) Find the magnitude of the tension T in the string. Use system of axes x-y as shown to write all forces in their Full Document, PHYS1A_2016_Winter_Discussion,_Week_04_Midterm_review, Physics 1A Fall 2018 midterm solutions.pdf, University of California, Los Angeles • PHYSICS 1A, University of California, Los Angeles • Physics 1A, Copyright © 2020. Let the small blue point be the box The frictional force then will adjust itself in precisely the right way to cancel any forces pulling Using this convention for your - and -axes, draw a force diagram on the block of mass 3 0 obj
It is as if the block is simply in f mg sin( θ) θ y 90 - θ Inclined Plane Problems push/pull Instead of an x-y coordinate system, inclined plane problems use a _____ coordinate system. the slope. Fa = (30 , 0) M = 100 Kg, g = 10 m/s^2 Free Body Diagram Consider the previous problem. f mg sin( θ) θ y 90 - θ Will the system accelerate uphill, downhill, or not at all? 1. Is your value for the acceleration in Part iii. An inclined plane is basically a ramp. What is Fg Block from problem five has a mass of 54.3 kg. It is as if the block is simply 0, sin Solutioneval(ez_write_tag([[728,90],'problemsphysics_com-medrectangle-4','ezslot_2',341,'0','0'])); 0000001443 00000 n
W weight of the box, N the normal force exerted by the inclined plane on the box, Fa the applied force so that the box is lowered at constant speed and Fk kinetic force of friction (box is moving downward, force of friction opposite motion). endobj
force on 12. (Neglect air resistance, but include friction between the block and the wedge.) Answer: unknowns: and . What is the acceleration of the block? substitute |Fk| by μk M g cos α into equation (1) Interpret the meaning of the expression you have just derived in Part ii. where, of course, 9. �J���˚A�m19���|] >�v?���]�"��Π��r�)�3�6y���6m��z?��f�5�c# nR�ǃ�� sense? If cos satisfies this condition, the system will accelerate uphill. 2 0 obj
Sure we can—use Newton’s 2nd law on ! If → 90 , then cos W + N = M a , M is the mass of the box |a| = M g sin (27°) / M = g sin (27) m/s^2 ≈ 4.5 m/s^2 weight: |W| = M g ; g = 10 m/ 45 33
a = (aeval(ez_write_tag([[300,250],'problemsphysics_com-box-4','ezslot_5',260,'0','0']));x , ay) = (|a| , 0) , box moving down the inclined plane in the direction of positive x hence ay = 0. Since the system is at rest, 0 (see Problem 11). Consider again the Inclined Plane Problem, this time we’ll include friction. If the coefficient of kinetic friction is 0.25, what is the acceleration of the block? |,�+ՠk5}D��T2�Ve|AH�(Ǯ:i�x����$W�z���3�Y���Jn�i'�澸Mz����HK�S����4|��j rvh�\���10� �eǺ�~o�~w���p7S�!p����Y��a'�VVi�?�(�@0����>86�P�U k!��T1i��28`̱�D��ĘЇ���gˎ=����,�n�̓�������,�sݡd���v��r�,hHړ���m���i� ��>�r�����=N�4C�i#8�6�aP�kF�?�>�/��. 0 . kinetic force of friction formula: |Fk| = μk |N| = μk M g cos α - M g sin α + |Fa| cos α + μs( |Fa| sin α + M g cos α ) = 0 ii. |T| cos (25°) = μs [ M g cos(35°) - |T| sin (25°) ] + M g sin(35°) A force F. of magnitude 30 N acts on the particle in the direction parallel and up the inclined plane. Since the system “wants” to move down the incline, the frictional force must be directed up the c) Find the magnitude of the force of friction acting on the particle. There is a _____ explanation of the why the angles above are both “theta”. W = (Wx , Wy) = ( - M g sin(35°) , - M g cos(35°)) component equations 2 0 obj
(Neglect air resistance, but include friction between the block and the wedge.) The Fa + W + N + Fs = 0 in components form: �'R�x�u��vh3�\goG����7j��q�K�w�t-V����2C��پ{������[*�������p�BLI���
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Draw a free-body diagram on the block of mass labeling all forces, and decompose each - M g cos α + |N| - |Fa| sin α + 0 = 0 (equation 2) slope. 3.85 / is positive, and it should be! endobj
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If cos is satisfies this condition, the system will accelerate downhill. |N| = 50 cos (30) = 25 √3 ≈ 43.3 N. A box of mass M = 10 Kg rests on a 35° inclined plane with the horizontal. 0000008329 00000 n
Inclined Plane Problems push/pull Instead of an x-y coordinate system, inclined plane problems use a _____ coordinate system. It is very possible that neither of these conditions for moving uphill or downhill are met. Answer: ii. Suppose the force of friction is strong enough to just barely enough stop the system from We now need to solve the system of two equations with two unknowns |T| and |N|. θ x x θ mg cos(θ) θ θ Fg = mg θ 6. What is the acceleration of the block? N = (0 , Ny) = (0 , |N|) x The string makes an angle of 25 ° with the inclined plane. A box of mass M = 10 Kg rests on a 35° inclined plane with the horizontal. 1 to satisfy Newton’s 2 law with zero acceleration. 0000029171 00000 n
|N| = |W| cos (27°) = 2 × 10 cos (27°) ≈ 17.8 N, Free Body Diagram endstream
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Suppose the force of friction is strong enough to stop the block from sliding. (For more on vectors, check out Chapter 3.) Inclined Plane (No Friction) H�T�=o�0�w~ō�:�ؤ !5I�~�I�sP�b,C����g���m=�a�~���}m�؇����:����
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