Chapter 4

Dynamics: Newton’s Laws of Motion

Types of Forces: An Overview

Examples of Nonfundamental Forces -- All of these are derived from the electroweak force:

normal or support forces

friction

tension in a rope

The Tension Force

Cables and ropes transmit forces through tension.

A force T is being applied to the right end of a rope.

For a massless rope, all of the force is transmitted to the box from the left end of the rope.

The box exerts an equal and opposite force to the left end of the rope via Newton’s 3rd law.

For a massless rope, F∑ =m

a = 0 ⇒ tension force the same on both ends

The Tension Force

A massless rope will transmit tension undiminished from one end to the other. If the rope passes around a massless, frictionless pulley, the tension will be transmitted to the other end of the rope undiminished.

Example. A supertanker of mass m = 1.50 x 108 kg is being towed by two tugboats. The tensions in the towing cables apply the forces T1 and T2 at equal angles of 30.0o with respect to the tanker’s axis. In addition, the tanker’s engines produce a forward drive force D, whose magnitude is D = 75.0 x 103 N. Moreover, the water applies an opposing force R, whose magnitude is R = 40.0 x 103 N. The tanker moves forward with an acceleration that points along the tanker’s axis and has a magnitude of 2.00 x 10-3 m/s2. Find the magnitudes of the tensions T1 and T2.

Application of Newton’s Laws of Motion

Application of Newton’s Laws of Motion

m = 1.50 x 108 kg R = 40.0 x 103 N D = 75.0 x 103 N ax = 2.00 x 10-3 m/s2 ay = 0

Application of Newton’s Laws of Motion

Force x component y component

1T

2T

D

R

0.30cos1T+

0.30cos2T+

0

0

D+

R−

0.30sin1T+

0.30sin2T−

Fy =∑ +T1 sin30.0 −T2 sin30.0 =may = 0 ⇒ T1 = T2 = T

Fx =∑ +T1 cos30.0 +T2 cos30.0+D− R =max

∴T = max + R−D2cos30.0

=1.50×108( ) 2.00×10−3( )+ 40.0×103 − 75.0×103

2 0.866( ) =1.53×105 N

Static and Kinetic Frictional Forces

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When an object is in contact with a surface there is a force acting on that object. The component of this force that is parallel to the surface is called the frictional force.

“cold welds”

Static and Kinetic Frictional Forces

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When the two surfaces are not sliding across one another the friction is called static friction.

Static and Kinetic Frictional Forces

The magnitude of the static frictional force can have any value from zero up to a maximum value.

MAXss ff ≤

NsMAXs Ff µ=

10 << sµ is called the coefficient of static friction.

Not a vector equation! fS is parallel to the surface, FN is perpendicular to the surface.

Static and Kinetic Frictional Forces

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Note that the magnitude of the frictional force does not depend on the contact area of the surfaces.

Static and Kinetic Frictional Forces

Static friction opposes the impending relative motion between two objects. Kinetic friction opposes the relative sliding motion that actually does occur.

Nkk Ff µ=

10 << sµ is called the coefficient of kinetic friction. k

Static and Kinetic Frictional Forces

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Usually, µs > µk

Static and Kinetic Frictional Forces

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Example. A sled and a rider are moving at a speed of 4.0 m/s along a horizontal stretch of snow. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050 and the mass of the sled and rider is 40 kg. Find the kinetic frictional force and the displacement, x, of the sled.

Static and Kinetic Frictional Forces

1.  Use Newton’s 2nd law in x and y directions. ΣFx = -fk = max --> ax = -fk/m = -µkFN/m (since fk = µkFN) ΣFy = FN - W = FN - mg = may = 0 --> FN = mg fk = µkFN = µkmg = (0.050)(40)(9.8) = 20 N

Static and Kinetic Frictional Forces

ax = -µkFN/m = -µkmg/m = -µkg = -(0.050)(9.8) = -0.49 m/s2 2. Solve for x using ax and kinematic equations. x v0x vx ax t ? 4.0 m/s 0 m/s -0.49 m/s2 vx

2 = v0x2 + 2axx --> x = (vx

2 - v0x2)/(2ax)

= (02 - 4.02)/(2(-0.49)) = 16 m

independent of mass of sled+rider

fk

Example. Block 1 (mass m1 = 8.00 kg) is moving on a 30.0o incline with a coefficient of kinetic friction between the block and incline of 0.300. This block is connected to block 2 (mass m2 = 22.0 kg) by a massless cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord.

fk fk

m1 = 8.00 kg m2 = 22.0 kg µk = 0.300 Find a, T and T’

Block 1: Σ  Fx = -fk - W1 sin 30.0o + T = m1a

Σ  Fy = FN - W1 cos 30.0o = 0 è FN = W1 cos 30.0o Block 2: Σ  Fy = T’ - W2 = m2(-a) We also know: T = T’ since the pulley and cord are massless fk = µkFN = µkm1g cos 30.0o = (0.300)(8.00)(9.80)(0.866) = 20.4 N

fk

Equations we’re left with to solve for a and T: -fk - W1 sin 30.0o + T = m1a T - W2 = -m2a è T = W2 - m2a Substituting for T in the first equation: -fk - W1 sin 30.0o + W2 - m2a = m1a a = (-fk - W1 sin 30.0o + W2)/(m1 + m2) = (-20.4 - (8.00)(9.80)(0.500) + (22.0)(9.80))/(8.00 + 22.0) = 5.20 m/s2 T = W2 - m2a = (22.0)(9.80) - (22.0)(5.20) = 101 N