Physics - Friction

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Friction

Table of Contents

Introduction

Problems

Answers

Introduction

Friction is a force parallel to the surface in the opposite direction of the motion of the object. The force of friction is greater for objects at rest(static) than in motion(kinetic). The equations are similar

F_fr = _kF_N and F_fr = _sF_N

Where F_N is the force of the object on the service perpendicular to the surface.

The static friction is greater than the kinetic friction because _s > _k between the same surfaces. The static friction is the force that must be overcome to allow the object to start moving. The kinetic friction is a force trying to slow down and stop an object in motion.

A very important point to remember, the force due to friction is always opposite the direction of the motion of the object NOT the Sum of Forces.

If an object is sliding or rolling down an incline plane at a constant velocity, the Tangent of the Angle on an inclined plane is equal to _k.

If an object just starts to slide or roll down an incline plane, the Tangent of the Angle on an inclined plane is equal to _s.

Table of Contents

Problems

1.	An object resist sliding down an incline until the angle reaches 18.6^o. What is the Coefficient of static friction? Answer to Problem #1.
2.	How far up the inclined plane can the object be pulled in 8.50 seconds if _k is 0.15? Answer to Problem #2.
3.	The _k is 0.25. After starting to slide, what is the acceleration and tension on the rope? Answer to Problem #3.
4.	Calculate the acceleration if the _k is 0.18. Answer to Problem #4.
5.	Calculate the acceleration if the _k is 0.18. Answer to Problem #5.
6.	A 50.0 kg block is slid up an inclined plane with an elevation of 15.0^o. The initial velocity is 5.36 m/s and the block stops after traveling 4.17 m. What is the coefficient of kinetic friction? Answer to Problem #6.
7.	The block is accelerated 2.56 m/s² and the coefficient of kinetic friction is 0.18. What is the force on the rope? Answer to Problem #7.
8.	The kinetic coefficient of friction is 0.26. What is the acceleration of the block? Answer to Problem #8.

Table of Contents

Answers

1.	An object resist sliding down an incline until the angle reaches 18.6^o. What is the Coefficient of static friction?
	F_P = F_w Sin 18.6^o F_N = F_W Cos 18.6^o F_fr = F_N = F_P F_P overcomes F_fr under this condition. F_w Sin 18.6^o= F_W Cos 18.6^o Solve for . F_w Sin 18.6^o = -------------------- = Tan 18.6^o = 0.337 F_W Cos 18.6^o

2.	How far up the inclined plane can the object be pulled in 8.50 seconds if _k is 0.15?
	F_{P rope} = 600. N x Cos 20.0^o = 564 N This is the force parallel to the inclined plane applied by the rope. F_{N rope} = 600. N x Sin 20.0^o = 205 N This is the force pulling mass away from inclined plane by the rope. F_{N mass} = 45.0 kg x 9.80 m s^-2 x Cos 25.0^o = 400. N This is the force normal to the plane applied by the mass. F_{P mass} = 45.0 kg x 9.80 m s^-2 x Sin 25.0^o = 186 N This is the force parallel and down the plane applied by the mass. F_N = 400. N - 205 N = 195 N F_N = F_{N mass} - F_{N rope} F_fr = F_N = 0.15 x 195 N = 29 N This force is parallel and down the inclined plane. F_parallel = F_{P rope} - F_{P mass} - F_fr Acceleration up the inclined plane is only dependent on Forces parallel to the inclined plane. F_parallel = 564 N - 186 N - 29 N = 349 N F_parallel 349 N F_parallel = ma thus a = --------------- = ------------ = 7.76 m s^-2 m 45.0 kg X = 1/2 a t² = 1/2 x 7.76 m s^-2 x (8.50 s)² = 280. m Initial velocity was zero.

3.	The _k is 0.25. After starting to slide, what is the acceleration and tension on the rope?
	F_{fr Red mass} = F_N = 0.25 x 25.0 kg x 9.80 m s^-2 = 61 N This force is in opposition to F_w blue mass. Fill in the equation F = ma for each mass. Down for blue mass is positive and to the right for red mass is positive. T - F_fr = m_red a m _blue g - T = m_blue a Solve each equation for T. T = F_fr + m_red a T = m _blue g - m_blue a T = T and F_fr + m_red a = m _blue g - m_blue a Solve for a. m_red a + m_blue a = m _blue g - F_fr m _blue g - F_fr 100 kg x 9.80 m s^-2 - 61 N a = -------------------- = ------------------------------------------- = 7.35 m s^-2 m_red + m_blue 25.0 kg + 100. kg Use either equation to solve for T. T = m _blue g - m_blue a = 100. kg (9.80 - 7.35) m s^-2 = 245 N

4.	Calculate the acceleration if the _k is 0.18.
	F_{N mass} = 50.0 kg x 9.80 m s^-2 x Cos 30.0^o = 424 N F_{P mass} = 50.0 kg x 9.80 m s^-2 x Sin 30.0^o = 245 N F_fr = F_N = 0.18 x 424 N = 76 N Since F_rope which is up and parallel to the inclined plane is greater than F_{P mass} the object will move up the inclined plane. F_fr is always opposite the direction of motion. F_parallel = F_{P rope} - F_{P mass} - F_fr F_parallel = 600 N - 245 N - 76 N = 279 N F_parallel 279 N F_parallel = ma thus a = --------------- = ------------ = 5.58 m s^-2 up the inclined plane m 50.0 kg

5.	Calculate the acceleration if the _k is 0.18.
	F_{N mass} = 50.0 kg x 9.80 m s^-2 x Cos 30.0^o = 424 N F_{P mass} = 50.0 kg x 9.80 m s^-2 x Sin 30.0^o = 245 N F_fr = F_N = 0.18 x 424 N = 76 N Since F_rope which is up and parallel to the inclined plane is less than F_{P mass} the object will move down the inclined plane. F_fr is always opposite the direction of motion. F_parallel = F_{P mass} - F_fr - F_{P rope} F_parallel = 245 N - 76 N - 100 N = 69 N F_parallel 69 N F_parallel = ma thus a = --------------- = ------------ = 1.38 m s^-2 down the inclined plane m 50.0 kg

6.	A 50.0 kg block is slid up an inclined plane with an elevation of 15.0^o. The initial velocity is 5.36 m/s and the block stops after traveling 4.17 m. What is the coefficient of kinetic friction?
	F_{N mass} = 50.0 kg x 9.80 m s^-2 x Cos 15.0^o = 473 N F_{P mass} = 50.0 kg x 9.80 m s^-2 x Sin 15.0^o = 127 N v² = v_o² + 2a(x - x_o) Solve for acceleration. v² - v_o² 0² - (5.36 m/s)² a = ------------- = ---------------------- = - 3.44 m s^-2 2(x - x_o) 2(4.17 m - 0 m) F_parallel = ma = 50.0 kg x (-3.44 m s^-2) = -172 N The acceleration is down the inclined plane. The F_fr is down the inclined plane since the object is moving up the inclined plane. We can change the down direction to the positive and change the sign on the Total force. We can now solve for F_fr. F_parallel = F_{P mass} + F_fr => F_fr = F_parallel - F_{P mass} = 172 N - 127 N = 45 N F_fr = F_N Solve for . F_fr 45 N = ------ = ------------ = 0.095 F_N 473 N

7.	The block is accelerated 2.56 m/s² and the coefficient of kinetic friction is 0.18. What is the force on the rope?
	F_{rope N} = F_rope x Sin 25.0^o F_{rope P} = F_rope x Cos 25.0^o F_parallel = ma = 60.0 kg x 2.56 m s^-2 = 154 N F_fr = F_{N mass} = 0.18 x 60.0 kg x 9.80 m s^-2 = 106 N F_fr is in the opposite direction of the motion. F_parallel = F_{rope P} - F_fr Solve for F_{rope P}. F_{rope P} = F_parallel + F_fr = 154 N + 106 N = 260. N F_{rope P} = F_rope x Cos 25.0^o Solve for F_rope. F_{rope P} 260. N F_rope = --------------- = --------------- = 287 N Cos 25.0^o Cos 25.0^o

8.	The kinetic coefficient of friction is 0.26. What is the acceleration of the block?
	F_{N mass} = 15.0 kg x 9.80 m s^-2 x Cos 28.0^o = 130. N F_{P mass} = 15.0 kg x 9.80 m s^-2 x Sin 28.0^o = 69.0 N F_fr = F_{N mass} = 0.26 x 130. N = 33.8 N Since F_{P mass} is greater than F_fr the object accelerates downward. If F_fr was greater, the object would not slide in either direction. F_parallel = F_{P mass} - F_fr = 69.0 N - 33.8 N = 35.2 N F_parallel = ma Solve for acceleration. F_parallel 35.2 N a = --------------- = ---------- = 2.35 m s^-2 down the inclined plane m 15.0 kg

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