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PHYS1110H, 2011 Fall. Shijie Zhong 2. Applications of Newton’s Laws - continued. . Contact force: Tension of a string, and friction Example 5. Block and String A block of a mass M on a friction-free surface is attached to a string with mass m. A force F is applied to the string to pull the block. What’s force that the string is applied to the block? Applying Newton’s 2nd law to both block and string: € € (2.14) F1 = Ma M , (2.15) F − F1 = ma m , where aM and am are the acceleration for the block and string, respectively, and are equal to each other . Combining these two equations, we obtain: a M = a m = F /(M + m) , (2.16) M F1 = F. (2.17) (M + m) € € Note 3 1 If m is zero or the mass of the string is ignored, F1=F. Although we often assume this in solving mechanics problems, as we see here, it is only an assumption or approximation to the reality. Example 6, Tension on a whirling rope. A uniform rope of mass M and length L is pivoted at one end and whirls with uniform angular velocity ω. Assuming that gravity can be ignored, what is the tension in the rope at distance r from the pivot? Consider a small segment of the rope between r and r+Δr. The segment has a length of Δr, and a mass of m=MΔr/L. Now let’s treat this segment as a particle and consider the forces and Newton’s law on this segment. The inward tensional force from the rope to the segment is T(r), and the outward force to the segment is T(r+Δr). This segment has an acceleration rω2 pointing towards the center or pivot. Suppose that the positive r direction is outward, then the acceleration is negative, and the NT’s 2nd law leads to: € € € € (2.18) T(r + Δr) − T(r) = −mrω 2 , M T(r + Δr) − T(r) = − Δr ⋅ rω 2 , (2.19) L T(r + Δr) − T(r) Mrω 2 =− . (2.20) Δr L Let Δr approach to zero, and use the definition of derivative, (2.20) becomes dT Mrω 2 =− , (2.21) dr L Note 3 2 which is a first order ODE that can be solved for how T depends on position r of the rope. Re-arrange (2.21) and integrate it. Mω 2 2 T =− r + C, (2.22) 2L € € € where C is an integration constant and can be determined by considering the constraint that tensional force T is zero at the tip of the rope, that is, T=0 at r=L, which leads to Mω 2 C= L, 2 Mω 2 2 2 T= (L − r ) . (2.23) 2L (2.23) describes how tensional force varies with r with the maximum tension at the pivot r=0, and zero tension at r=L. Example 7. Force on a pulley from a string. A string with constant tension T is deflected through angle 2θ0 by a smooth fixed pulley. What’s the force on the pulley? Due to the symmetry, the y component of the force on the pulley is zero, and we do not need to consider it. Consider a small segment of the deflected string that starts at θ and ends at θ+Δθ (see the figure). There are two tensional forces pulling on the segment from each end, but they are not parallel. There is also a force pushing the segment outward from the pulley ΔF. At the equilibrium, these three forces cancel. In the tangential direction, Note 3 3 the two tensional forces cancel each other. In the radial direction, or the direction of force ΔF, the force balance is (i.e., when acceleration is zero): ΔF − 2T sin(Δθ /2) = 0 , (2.24) For small x, sin(x) ≈ x . Therefore, sin(Δθ /2) = Δθ /2 , because Δθ/2 is small. € ΔF = 2TΔθ /2 = TΔθ . (2.25) € x component of this € The force ΔF on the pulley is € € ΔFx = ΔF cos θ = TΔθ cos θ . (2.26) The x component of the total force on the pulley is the sum or integral of ΔFx for all the small segments or for θ to vary from -θ0 to θ0. θ0 Fx = ∫ ΔFx = ∫ T cos θdθ = 2T sin θ 0 . (2.27) −θ 0 € We do not need to compute y component of the force on the pulley, because they all cancel. Friction. Friction arises when the surface of one body moves or tries to move along the surface of another body. For many surfaces, the maximum friction is equal to µsN, where µs is coefficient of static friction, and N is the normal force against the surface. When there is a relative motion between two bodies with a contact surface, then the friction force has a magnitude of µkN and a direction opposite to the motion, where µk is the coefficient of kinetic friction. For bodies not in relative motion, the frictional force is between 0 and µsN. For example, when a force F is applied to a body but is not sufficient to overcome the maximum frictional force µsN (i.e., F< µsN), then the frictional force is equal to F in magnitude and is opposite in direction (see the figure). Static friction coefficient µs is larger than kinetic friction coefficient µk. Note 3 4 Example 8. The spinning terror. A large vertical cylinder has a coefficient of friction µ at its inner wall, and has a radius of R. How fast does the cylinder must rotate to hold an object to the wall with no support at its bottom? Suppose that the angular velocity is ω, the normal force to an object with mass m from the wall is N = mRω 2 , and points to the center axis of the cylinder. The maximum frictional force on the object is f=µN, and the direction of the frictional force is upwards, opposite to the direction of the gravitational force. To hold the object to the wall and to prevent the object € from sliding down the wall, the frictional force f must be equal to the weight of the object, mg. f = µN = µmRω 2 = mg , or µRω 2 = g . Therefore, the minimum angular velocity must satisfy: € € ω 2 ≥ g /(µR) . € (2.28) The larger the radius R, or the larger the coefficient of friction, the slower the cylinder can rotate to hold the object onto the wall. Example 9. Air drag. v0 , suppose that there is an air For a cannon ball firing at an initial velocity f drag force that is given as = −mkv , where v is velocity vector, m is the Note 3 € € € 5 mass of the cannon ball, and k is a constant. What’s velocity as a function of time during the flight? € € € € € Fnet = −mgeˆ y − mkv = −(mg + mkv y )eˆ y − mkv x eˆ x , (2.29) dv y dv −kv x eˆ x − (mg + mkv y )eˆ y = m( x eˆ x + eˆ y ) , (2.30) dt dt (2.30) is the NT’s 2nd law. Treat each component separately and we have dv −kmv x = m x , (2.31) dt dv y −mg − mkv y = m , (2.32) dt We have solved these two equations (2.31) and (2.32) in our homework set 2, Q3. The solutions are given as follows.. g g (2.33) v x = v x 0 e−kt , v y = (v y 0 + )e−kt − . k k The air drag causes the horizontal velocity of the ball to decrease with time, and also a terminal velocity for vertical component, -g/k. € Note 3 6