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Transcript
Tension Tension • Tension is a pulling force that arises when a rope, string, or other long thin material resists being pulled apart without stretching significantly. • Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string. A physical picture of tension Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string. Tension examples Note that the pulleys shown are magic! They don’t affect the tension in any way, and serve only to bend the line of action of the force. Sample problem A. B. A 1,500 kg crate hangs motionless from a crane cable. What is the tension in the cable? Ignore the mass of the cable. Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now? Springs (Hooke’s Law) • The magnitude of the force exerted by a spring is proportional to the amount it is stretched. • F = kx – F: force exerted by the spring (N) – k: force constant of the spring (N/m or N/cm) – x: displacement from equilibrium (unstretched and uncompressed) position (m or cm) • The direction of the force is back toward the equilibrium (or unstretched) position. Sample problem • A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length? Sample problem • A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60. Laboratory • Using the ramp at an angle, determine the coefficient of kinetic friction between the felt side of the block and the ramp by allowing the block to accelerate down the ramp. Tomorrow, you will turn in the appropriate free body diagrams and calculations. Sample problem A 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown, A) what is the acceleration of the system? B) what is the tension in the string connecting the objects? (Assume a frictionless surface.) Gravity A very common accelerating force is gravity. Here is gravity in action. The acceleration is g. Slowing gravity down The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g. Magic pulleys on a flat table • Magic pulleys bend the line of action of the force without affecting tension. N m1g -x T SF = ma m2g + T – T = (m1 + m2)a a = m2g/(m1+m2) m1 T Frictionless table m2 x m2g Sample problem Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find (a)the acceleration of each block. (b)the tension in the connecting string. m1 m2 Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary. m1 m2 Sample problem - solution fs m1g N SF = 0 m2g - T + T – fs = 0 fs = m2g msN = m2g msm1g = m2g ms = m2/m1 = 0.50 T m1 T m2 m2g Note: we know from previous problem that the static friction is not enough to hold the blocks in place! Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If ms = 0.30 and mk = 0.20, what is (a)the acceleration of each block? (b)the tension in the connecting string? m1 m2 Sample problem – solution (a) fk m1g N SF = ma m2g - T + T – fk = ma m2g - mkm1g = (m1 + m2)a a = (m2- mkm1) g/(m1 + m2) a = 2.0 m/s2 T m1 T m2 m2g Using that the acceleration is 2.0 m/s2 from part a) Sample problem – solution (b) fk m1g N Using block 2 SF = ma m2g - T = m2a T = m2(g – a) T = 40 N Using block 1 SF = ma T - fk = m1a T = m1(a + mkg) T = 40 N T m1 T m2 m2g Magic pulleys on a ramp • It’s a little more complicated when a magic pulley is installed on a ramp. N SF = ma m2g -T + T – m1gsinq = (m1+m2)a T m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) T m1gsinq m1g m1gcosq m2 q m2g Sample problem • Two blocks are connected by a string as shown in the figure. What is the acceleration, assuming there is no friction? 5 kg 45o Sample problem - solution N T m1gsinq m1g m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a T a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = - 1.35 m/s2 5 kg m2g 45o Sample problem - solution N T SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = - 1.35 m/s2 T m1gsinq m1g m1gcosq 5 kg m2g 45o How would this change if there is friction on the ramp?