Principles of Physics Chapter 2 PDF

Principles of Physics Dr. Mohaned Mohmmed School of Biotechnology, Badr University in Assiut  Chapter Two Statics and Torque Physics Dr. Mohaned Mohammed 2 Torque The magnitude of torque τ equals the applied force F times the length of arm r that is perp...

Principles of Physics Dr. Mohaned Mohmmed School of Biotechnology, Badr University in Assiut  Chapter Two Statics and Torque Physics Dr. Mohaned Mohammed 2 Torque The magnitude of torque τ equals the applied force F times the length of arm r that is perpendicular to the applied force. 𝜏ҧ = 𝐹 × 𝑟 = 𝑟𝐹 sin 𝜃 where θ is the angle between the direction of the force and the direction of r. τ Torque is zero if the force is passing through the pivot point. F Force is measured in N (Newton) and the torque is measured in N.m. The direction of the torque can be obtained by the right-hand rule. Physics Dr. Mohaned Mohammed 3 Physics Dr. Mohaned Mohammed 4 Static Equilibrium A body is said to be in mechanical equilibrium if the next two conditions are verified: 1. The resultant external force must equal zero ΣF = 0 2. The resultant external torque about any axis must equal zero Στ=0 otherwise, the body will rotate. A body is said to be static if the linear and angular velocities are zero. Physics Dr. Mohaned Mohammed 5 Conditions for Static Equilibrium The center of mass in the object is the average position of all object parts, according to their masses. If the gravity is constant all over the object parts, the center of mass can be considered as the center of gravity. In general, a body in stable equilibrium under the action of gravity has its center of the mass position directly over its base of support. c.g. c.g. c.g. Physics Stable Stable Dr. Mohaned Mohammed Unstable 6 Conditions for Static Equilibrium For the stable body, the reaction force Fr (upwards) cancels the force produced by the body weight Fw (downwards) as they work in the same line with opposite directions. When the body is unstable, i.e., rotates around a pivoting point, there are two conditions: Physics Dr. Mohaned Mohammed 7 Physics Dr. Mohaned Mohammed 8  Stable Unstable Unstable Unstable Physics Dr. Mohaned Mohammed 9 Conditions for Static Equilibrium If its center of mass is still above its base, the reaction force at the pivoting point and force produced by the body weight try to restore the body to its original position. If the center of mass is outside the base, the produced torque tends to topple the body. Physics Dr. Mohaned Mohammed 10 Equilibrium in the Human Body The center of gravity of an erect person with arms at the side is at approximately 56% of the person’s height measured from the soles of the feet. Note: The center of mass changes its position if the shape of the body changes, even if, the mass is kept constant. Physics Dr. Mohaned Mohammed 11  Equilibrium in the Human Body When carrying an uneven load, the body tends to compensate by bending and extending the limbs to shift the center of gravity back over the feet. People who have lost an arm often have some problems (i.e., permanent distortion of the spine), because of the continuous compensatory bending of the torso. Therefore, the amputees should wear an artificial arm to restore balanced weight distribution. Physics Dr. Mohaned Mohammed 12 The Action of an External Force on Human Body Stability Example 1: Let us assume a person as shown in Fig. Calculate the force applied to the shoulder to topple a person standing at rigid attention (Comparing the torques about point A). Assuming that the mass of the person is 70 kg (the gravitational acceleration g= 9.8 m/s2) Physics Dr. Mohaned Mohammed 13 Answer of Example 1 The anti-clockwise torque τa is produced by the applied force Fa by pivoting around point A is τa = Fa x ra = Fa x 1.5 The opposite restoring torque τw due to the person’s weight W is τw = Fw x rw = 70 x 9.8 x 0.1 = 68.6 N.m The person is on the threshold of falling when the magnitudes of these two torques are equal τa = τw Fa x 1.5 = 68.6 The force required to topple an erect person is larger than Fa = 68.6/1.5 = 45.7 N Physics Dr. Mohaned Mohammed 14 Example 2 A person is standing at rigid attention with mass of 70 kg. Its height measured from his shoulder is 150 cm and his foot width is 10 cm. Calculate the magnitude of the applied external force required to topple this person. Assume the gravitational acceleration to be 10 m/s2. Physics Dr. Mohaned Mohammed 15 Answer of Example 2 The anti-clockwise torque τa is produced by the applied force Fa by pivoting around point A is τa = Fa x ra = Fa x 1.2 The opposite restoring torque τw due to the person’s weight W is τw = Fw x rw = 70 x 9.8 x 0.55 = 377.3 N.m The person is on the threshold of falling when the magnitudes of these two torques are equal τa = τw Fa x 1.2 = 377.3 The force required to topple an erect person is larger than Fa = 377.3/1.2 = 314.42 N Physics Dr. Mohaned Mohammed 16 Example 3 For the situation shown in the following figure, calculate: (a) FR, the force exerted by the right hand, and (b) FL, the force exerted by the left hand. The hands are 0.9 m apart, and the cg of the pole is 0.6 m from the left hand. The weight of the bar is 5 Kg. Physics Dr. Mohaned Mohammed 17 Answer of Example 3 Solution for (a) There are now only two nonzero torques, those from the gravitational force (τw) and from the push or pull of the right hand (τR). Stating the second condition in terms of clockwise and counterclockwise torques, or the algebraic sum of the torques is zero. Here this is τR = - τw since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise torque. Using the definition of torque, τ = rF sinθ, noting that θ = 90o, and substituting known values, we obtain 0.9 x FR = 0.6 X mg = 0.6 x 5 x 9.8 Thus, FR = 32.7 N Physics Dr. Mohaned Mohammed 18 Answer of Example 3 Solution for (b) The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law: FL + FR – mg = 0 From this we can conclude: FL + FR = mg Solving for FL, we obtain FL = 5 x 9.8 – 32.7 = 16.3 N Physics Dr. Mohaned Mohammed 19  Homework For the situation shown in the following figure, calculate: (a) FR, the force exerted by the right hand, and (b) FL, the force exerted by the left hand. The hands are 0.9 m apart, and the cg of the pole is 0.6 m from the left hand. The weight of the bar is 5 Kg. Physics Dr. Mohaned Mohammed 20  Muscles The muscles producing skeletal movements consist of thousands of parallel fibers. The force exerted by the muscle depends on the number of fibers contracting the muscle. The thousands of fibers give a high variability of the force. The tendons attach the muscle to the bones. Most muscles end in a single tendon. Except for biceps and triceps, they end in two and three tendons, respectively, and each tendon is attached Physics to a different bone. Dr. Mohaned Mohammed 21 Levers A lever is a rigid bar free to rotate about a fixed point called the fulcrum. The position of the fulcrum is fixed with respect to the bar. Levers are used for lifting loads in an advantageous way and to transfer movement from one point to another. There are three parts to all levers: Fulcrum - the point at which the lever rotates. Input force (also called the effort) - the force applied to the lever. Output force (also called the load) - the force applied by the lever to move the load. Physics Dr. Mohaned Mohammed 22 Levers If the lever is in equilibrium, then, the torque produced by the applied force F is equal to the torque produced by the load W. If the load and force arms are d1 and d2, respectively then F d2 = W d1 By defining the mechanical advantage of the lever as the ratio of the weight to the applied force, we get 𝑊 𝑑2 𝑀= = 𝐹 𝑑1 Physics Dr. Mohaned Mohammed 23  Levers There are three classes of levers as shown in the figure. In a Class 1 lever, the fulcrum is located between the applied force and the load. So, with d1 much smaller than d2, a large mechanical advantage M can be obtained. In a Class 2 lever, the load is located between the applied force and the fulcrum. Here, d1 is always smaller than d2 so the mechanical advantage is M >1. In a Class 3 lever, the applied force is located between the load and the fulcrum. d1 is always larger than d2 so the mechanical Physics advantage is M