Mech-ing

Question: Consider material aluminum and area of beam is 1in × 1in. To determine deflection of beam with F A  = 87.5lbf and F B = 47lbf. The length of beam is 60 inches. F A  is applied at 15 inches from one support and F B  is applied at 45 inches from this same support. Determine deflection of beam. Answer:

Question : A car starts from rest and goes down a slope with a constant acceleration of 5 m/s^2. After 5 seconds, the car reaches the bottom of the hill. What is its speed at the bottom Of the hill, in meters per second? Answer: Given data: Car starts from rest so the initial velocity of tge car (u) = 0 m/s Acceleration of the car (a) = 5 m/s² Time required to reach at bottom (t) = 5 seconds According to the uniform acceleration motion, v = u + at Here, v is the velocity of the car at the bottom v = 0 + 5(5) v= 25 m/s Thus, the velocity of the car at the bottom is 25 m/s.

Question: When an archer shoots an arrow straight up with an initial velocity magnitude of 100.0 m/s. After 5.00 s the velocity is 51.0 m/s. At what rate is the arrow decelerated? Answer: Given data: Initial velocity of the arrow (u) = 100 m/s, Travelling time (t) = 5 seconds, Final velocity of the arrow (v) = 51 m/s According to the equation of the constant acceleration, acceleration can be written as,   Thus, the acceleration of the arrow is 9.8 m/s.

Question: A car accelerates uniformly at a rate of 10m/s² from an initial velocity of 36km/hr for 30s. What is the covered distance during this period? Answer: Given data: Acceleration of the car (a) = 10 m/s², Initial velocity of the car (u) = 36 km/h Time of travelling (t) = 30 s Step-1 To covert velocity in m/s from km/h. As we know that 1 km/h = 1000/3600 m/s Therefore, Step-2 To determine distance travelled by the car within t = 30 s. According to the equation of the constant acceleration motion. Distance travelled by the car can be written as,   Thus, the car will travel through 4800 m within 30 seconds.

Question: A car has an acceleration of 6 m/s². And it is started from the rest. How long will it take to reach a velocity of 48 m/s? Answer: Given data: Acceleration of the car (a) = 6 m/s²,  Initial velocity of the car (u) = 0 m/s (because intially car is at rest) and  Final velocity of the car (v) = 48 m/s. According to the equation of the constant acceleration motion,   Thus, the time required to obtain 48 m/s velocity is 8 seconds.

Question: How much centripetal force is needed to keep a 0.24 kg ball on a 1.72 m string moving in a circular path with a speed of 3.0 m/s? Answer:  Given data: Mass of the ball (m) = 0.24 kg, Length of the string (l) = 1.72 m, Speed of the ball (v) = 3 m/s As per definition of the centripetal force, mathematically, Centripetal force can be written as, Thus, the required centripetal force is 1.256 newton.

Question: A well insulated piston-cylinder system contains 5 liter of saturated liquid water. The piston has a diameter and mass of 0.30 m and 15 kg respectively. The system is placed at an area where the atmospheric pressure is at 86 kPa. The water is stirred by a paddle wheel while a current of 10A flows for 45 min through a resistor placed in the water. (a) Determine the initial pressure of the water (b) Determine the initial mass of the water in the cylinder. (c) If one -half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 300 kJ, determine the voltage of the source. (d)Determine the final volume of the water. Answer: Given data:  The initial volume , The atmospheric pressure  and mass of the piston  . (a) Initial pressure (b) From property table of water, specific volume of saturated water at 88.082 kPa is, and therefore mass of the water is, (c) Quality or dryness fraction (x) = 0.5 and from property ...

Question: State in words Newton's second law of motion assuming mass is not a constant. Write down the general equation of motion assuming mass is not a constant. Answer: The Second law of thermodynamic states that "the rate of change of momentum of a particle over time is directly proportional to the force applied and it is in same direction as the applied force". Mathematically, Here,   is the momentum of particle, and   is the applied force to the particle. Thus,  Here,   is the mass of the particle and   is the velocity of the particle. Here,   is the acceleration of the particle and   is the rate of change of the mass. Thus, the above equation represents the general equation of the motion. From this equation, it can be said that the applied force accelerates the particle and also changes the mass with respect to time.