T y lying flat on a horizontal surface, as we would expect. 1. 0 - 50 cos (30) + |N| + 0 = 0 nd If the force down the ramp is smaller than the maximum static friction, sin , and 0. the block will not slide at all, so The value of the frictional force, , will then exactly match the force down the ramp:

|F| = - 50 sin(30) + 30 = 5 N

Inclined Plane with Friction then surely you must have made a mistake!!! Can we solve for the tension ? 3) where μ is the coefficient of friction. |T| cos (25°) = μs [ M g cos(35°) - |T| sin (25°) ] + M g sin(35°) b) Find the magnitude of the tension T in the string. For exam-ple, take a look at the cart in Figure 5-1; it’s on an inclined plane, ready to roll. VECTOR ADDITION. W = (Wx , Wy) = (|W| cos (27°) , - |W| sin (27°)) Hence (For now, assume no friction in this problem. (For more on vectors, check out Chapter 3.) Algebra Review PDF.

The string makes an angle of 25 ° with the inclined plane. What is Define the -direction parallel and the -direction perpendicular to the slope. Newton’s 2nd law in the -direction once the system is accelerating gives

1 0 obj Draw the full free-body diagram of a block that is getting pushed DOWN an

W + N = M a , M is the mass of the box Interpret the meaning of the expression you have just derived in Part ii.

sin 3.85 , and .

0, and ��qm%�n��V3� ҡ����NP��+�Vd���T��$���4w#>7��j��QDΠK�`hn�'|���o �A83#�2R�`�8��Lp%�'��1�Ȟ���Ji��Dj�*M��x�� �=cġ�S��%FꩍL�t�S�Te�����,�Hĺ�[7:�vkl�C�O�-��P�5*`�[b���|*���ceu��j�"�M������,U��{� FY�܆���.W ���Z��g5vdj�2�h)�)S��jJ�Uy���Ul��3_y��)R~Q��"�J>x\%���SeA"�*4����:R#6�-k��Lʗ *��&|�R`���R�ձ�f��Bڧ�P��5%�s��+-��%|um(���ev]o�$����V��[_o_ȋ�a+o��[]$�p:HK�9x�YE�b���D f��v���X��A\cn!�M>������2㰓�ҵ�����M.c62!P�~�ؗ&I�:���V�^�Oi�˻eX�%�n?o�~Z��i! Free body A particle of mass 5 Kg rests on a 30° inclined plane with the horizontal.

N = (0 , Ny) = (0 , |N|) An inclined plane is basically a ramp. accelerate the system downhill is:

If the coefficient of kinetic friction is 0.25, what is the acceleration of the block?

The static coefficient of friction between the box and the inclined plane is μs = 0.3. law in the -direction reads: Suppose the system is just barely about to accelerate down the slope—it isn’t moving yet;

If cos satisfies this condition, the system will accelerate uphill. i. ii. Based off your answers to Problem 3, 0000015237 00000 n

this time, the system will accelerate in the opposite just about to. 1 15. x components: 0 + |Fa| + |Fk| - M g sin α = 0 (eq 1) ( Newton's second law) Include: normal force, friction, components of the gravitational force (mg) S4P-1-6 Calculate the components of exerted on an object resting on an inclined plane. endstream endobj 56 0 obj <> endobj 57 0 obj <>stream Write down Newton’s 2nd law in the -direction. |N| = M g cos α Using this convention for your - and -axes, draw a force diagram on the block of mass 14. endobj

Solve for the mass, c) Find the magnitude of the force of friction acting on the particle. 0000006983 00000 n 1, sin

T = (Tx , Ty) = (|T| cos (25°) , |T| sin (25°) ) sin α = 2/4 = 1/2

0000000016 00000 n Again, be careful! <> Answer: iii. . b), The box is at rest, hence its acceleration is equal to 0, therefore the sum of all forces acting on the box is equal to its mass times its acceleration which is zero. Two forces act on the box: the weight W of the box and N the force normal to and exerted by the inclined plane on the box (blue point) Substitute |N| by M g cos(35°) - |T| sin (25°) in eq 1 to get 77 0 obj <>stream A box of mass M = 10 Kg rests on a 35° inclined plane with the horizontal. 0000002843 00000 n |,�+ՠk5}D��T2�Ve|AH�(Ǯ:i�x����$W�z���3�Y���Jn�i'�澸Mz����HK�S����4|��j rvh�\���10� �eǺ�~o�~w���p7S�!p����Y��a'�VVi�?�(�@0����>86�P�U k!��T1i��28`̱�D��ĘЇ���gˎ=����,�n�̓�������,�sݡd���v��r�,hHړ���m���i� ��>�r�����=N�4C�i#8�6�aP�kF�?�>�/��. 0000029373 00000 n If the force down the ramp is equal to the maximum static friction, separately for the various cases. 5 | P a g e This gives me the value of 8. "i4�t� Ib1�K,�R1C���4�����ĳ��,�O�����/���IY�>�M���xr�����Oċc�����1��99 xr�$o�AS��ɥ#9�L?_?mw��ͭx��w��f�ԛj)�ڔ����m�k�'�*y�~&>:��ʞ8�a#�wu'z��3QI�ɚ���%��ɱ� ʁ�z! Terms. M = 100 Kg, g = 10 m/s^2 Fa = (30 , 0) Thus, to transform the problem sin 2 cos 0.5 0.433 Case 1 is satisfied!!! Answer: unknowns: and . 0000028905 00000 n 3) where μ is the coefficient of friction. is the case, then clearly, the acceleration will be zero: |N| = |W| cos (27°) = 2 × 10 cos (27°) ≈ 17.8 N, Free Body Diagram m2 Fg = m1g Fg = m2g 11. ;2�� in the -direction gives .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: x Inclined Plane Problems push/pull Instead of an x-y coordinate system, inclined plane problems use a _____ coordinate system. Since it isn’t moving yet, the acceleration must be zero, so Newton’s 2nd → 90 . 0000001443 00000 n 1. The box is the small blue point. endobj Interpret the meaning of the expression you have just derived in Part ii.

Solving for Find the magnitude of force Fa.

Fk = (|Fk| , 0) i. Forces represented by their components |Fa| = M g sin α - |Fk| = M g sin α - μk M g cos α Fs = (- |Fs| , 0) = ( - μs |N| , 0) , where μs is the coefficient of friction between the box and the inclined plane. 0000005243 00000 n (|W| sin (27°) , - |W| cos (27°)) + (0 , |N|) = M (|a| , 0) the -direction is aligned parallel to the slope and the -direction is aligned perpendicular to 1. If → 90 , then cos Since it isn’t moving yet, the acceleration must be zero, so Newton’s 2nd force on 12. m1 m1g sin( θ) f = ??? In the diagram below, W is the weight of the box, N the normal force exerted by the inclined plane on the box, Feval(ez_write_tag([[336,280],'problemsphysics_com-banner-1','ezslot_7',360,'0','0']));a is the force applied to have the box in equilibrium and Fs the force of friction opposite Fa. 16. solve equation 2 above for |N| to get T tension of string, W weight of the box, N force normal to and exerted by the inclined plane on the box, Fs is the force of friction, Equilibrium: W + T + N + Fs = 0 All rights reserved. Here are the 12.3 Notice that Privacy is just a tiny bit larger?

The system will therefore accelerate uphill. When solving problems about objects on an incline, it is convenient to choose a coordinate system with axes parallel and perpendicular to the surface as shown in Fig. Worksheet SOLUTIONS -Inclined Planes and Friction.pdf - WorksheetSOLUTIONS 1 Inclined Plane(No Friction To analyze this common problem it is first, 2 out of 2 people found this document helpful. Solving for

| = 70 [ sin 25° - 0.3 cos 25° ] / [ cos 25° + 0.3 sin 25° ] ≈ 10.2 N. Fg Block from problem five has a mass of 54.3 kg. 0000001778 00000 n ;Ǉ�J��;��x�1=鯥��\b���[��n��9FVt��\ �?K���2���/܈捯�65Zwqf����[�o�iY�Fʧ�B��U^k����fq��yΠ$[i�co�*���Έ �FBH�է2��TL�2��5 ��YT���� |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. sense? %PDF-1.5 5.

UTC Physics 1030L: Friction, Work, and the Inclined Plane 40 The magnitude of the frictional force, Ff, on an object, can also be described by: Ff = μN (eq. Sum of x components = 0 Suppose the force of friction is strong enough to stop the block from sliding. |T| cos (25°) = μs |N| + M g sin(35°)

0000003009 00000 n the system uphill is:

incline to oppose this motion and it will have reached the maximum value with the static x-components are equal : |W| sin (27°) + 0 = M |a|

0000003276 00000 n What will slope.

Solve for the mass,

Now you have to be very careful, for once the system is moving we must use kinetic friction in For two vectors to be equal, their components must be equal. θ x x θ mg cos(θ) θ θ Fg = mg θ 2. 2 �J���˚A�m19���|] >�v?���]�"��Π��r�)�3�6y���6m��z?��f�5�c# nR�ǃ�� sin ii.

��yIa���hd���E��*f��xݯ�Z�|��������G\�sq��BZW�uy�-dpS���n��j2W(��]����E��!��i[ܡ���7w��8���x\����M����as����l�wcY����v��E�����7�����Q\��ˎ����K[��ߖJ/~Z2�ܶ��.~#��l&U� cos This gives me the value of c) Find the magnitude of the force of friction acting on the particle. <> 0000003525 00000 n <>/XObject<>/ExtGState<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 612 792] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>>

i. Since the 0, the tension must be 4 (see Problem 11). 0.4 . A video with examples on Components of Vectors may be helpful. Unformatted text preview: Worksheet SOLUTIONS INCLINED PLANES AND FRICTION 1. - M g sin(35°) + |T| cos (25°) + 0 - μs |N| = 0 Fs = (-|F| , 0) component equations of the box on the inclined plane and label all forces acting on the box.

0000003975 00000 n 0000005539 00000 n Answer: , 2 , 30 , 0.5, 0.4 Will the system accelerate uphill, downhill, or not at all?

Answer: What is the tension on the string? θ θ m1g cos(θ) T

substitute |Fk| by μk M g cos α into equation (1) What is the value of the frictional force?

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