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happyphysics.com Physics Lecture Resources Prof. Mineesh Gulati Head-Physics Wing Happy Model Hr. Sec. School, Udhampur, J&K Website: happyphysics.com
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Ch 5 Appling Newton’s Law
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5.1 Using Newton’s First Law: Particles in Equilibrium Body is in equilibrium when it is at rest or moving with constant velocity in an inertial frame of reference. Body is in equilibrium when it is at rest or moving with constant velocity in an inertial frame of reference. Vector sum of all forces must be zero All component forces must be zero
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Example 5.3 Find the tension of three chains, assuming that w is given and the weights of the ring and chains are negligible. Find the tension of three chains, assuming that w is given and the weights of the ring and chains are negligible. ANS:
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Example 5.4 Find the tension in the cable and the force with which tracks push on the car’s tires. Find the tension in the cable and the force with which tracks push on the car’s tires.
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ANS:
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5.2 Using Newton’s second law: Dynamics of particles We start to consider the net force on the body is not zero, but is equal to the mass of body times its acceleration We start to consider the net force on the body is not zero, but is equal to the mass of body times its acceleration Newton’s second law: Component form:
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Example 5.9 Find the tension T in the supporting cable while the elevator is being brought to rest. Given the mass of elevator and load is 800kg, It moving downward at 10m/s and the distance for constant acceleration to stop is 25m. Find the tension T in the supporting cable while the elevator is being brought to rest. Given the mass of elevator and load is 800kg, It moving downward at 10m/s and the distance for constant acceleration to stop is 25m.
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ANS To determine a y, we need to used the constant acceleration equation
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5.3 Frictional Forces θ Mg sinθ Mg cosθ Fr
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Static friction force Kinetic friction force
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F increase f increase Box at rest No relative motion Box start to move and fs f k Relative motion
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Example 5.19 Given the mass of car is 12000N. If coefficient of rolling friction is 0.015, what is the horizontal force make the car move with constant speed on level road. Given the mass of car is 12000N. If coefficient of rolling friction is 0.015, what is the horizontal force make the car move with constant speed on level road. ANS:
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Fluid Resistance and Terminal Speed Direction of the fluid resistance force acting on a body is always opposite the direction of the body’s velocity to the fluid Direction of the fluid resistance force acting on a body is always opposite the direction of the body’s velocity to the fluid Fluid resistance at low speed: Fluid resistance at low speed: Fluid resistance at high speed: Fluid resistance at high speed: Depend on the shape and size of object
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Terminal Speed When the speed of the object increase, the resisting force also increase, until finally it is equal in magnitude to the weight. At this time the acceleration become zero and speed will be constant, This final speed v t called terminal speed. When the speed of the object increase, the resisting force also increase, until finally it is equal in magnitude to the weight. At this time the acceleration become zero and speed will be constant, This final speed v t called terminal speed.
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For f=kv, For f=kv, For f=Dv 2, For f=Dv 2,
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5.4 Dynamics of Circular Motion
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Summary
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END Visit: happyphysics.com For Physics Resources
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