1) A body of mass 200 kg is placed in a lift. Find the reaction between the floor of the lift and the body when the lift moves upward:
i) At constant velocity
ii) with an acceleration of [tex]$4.5 m / s ^2(g=10 m / g )$[/tex]
Two masses are connected by a light string which passes over a fixed smooth frictionless pulley.
Find the common acceleration [tex]$A$[/tex] and the tension [tex]$T$[/tex] in the string when the system is moving freely.
Answer (2)
Calculate the reaction force at constant velocity: R = m g = 200 × 10 = 2000 N.
Calculate the reaction force with acceleration: R = m ( g + a ) = 200 × ( 10 + 4.5 ) = 2900 N.
Calculate the common acceleration for the pulley system: A = m 1 + m 2 m 1 − m 2 g = 10 + 6 10 − 6 × 10 = 2.5 m/s^2.
Calculate the tension in the string: T = m 1 + m 2 2 m 1 m 2 g = 10 + 6 2 × 10 × 6 × 10 = 75 N.
Explanation
Problem Analysis We are given a problem involving a body in a lift and two masses connected by a string over a pulley. We need to find the reaction force between the floor of the lift and the body under different conditions, as well as the common acceleration and tension in the string for the pulley system.
Reaction Force at Constant Velocity In the first scenario, the lift moves upward at a constant velocity. This means there is no net acceleration. The reaction force R is equal to the weight of the body, which is given by R = m g , where m = 200 kg and g = 10 m / s 2 . Therefore, R = 200 k g × 10 m / s 2 = 2000 N .
Reaction Force with Acceleration In the second scenario, the lift moves upward with an acceleration of 4.5 m / s 2 . The net force on the body is ma . The reaction force R is given by R − m g = ma , so R = m ( g + a ) . Substituting the given values, we have R = 200 k g × ( 10 m / s 2 + 4.5 m / s 2 ) = 200 k g × 14.5 m / s 2 = 2900 N .
Common Acceleration For the pulley system, let m 1 = 10 kg and m 2 = 6 kg. The common acceleration A is given by A = m 1 + m 2 m 1 − m 2 g = 10 + 6 10 − 6 × 10 m / s 2 = 16 4 × 10 m / s 2 = 2.5 m / s 2 .
Tension in the String The tension T in the string is given by T = m 1 + m 2 2 m 1 m 2 g = 10 k g + 6 k g 2 × 10 k g × 6 k g × 10 m / s 2 = 16 120 × 10 N = 75 N .
Final Answer The reaction force when the lift moves upward at constant velocity is 2000 N. The reaction force when the lift moves upward with an acceleration of 4.5 m / s 2 is 2900 N. The common acceleration for the pulley system is 2.5 m / s 2 , and the tension in the string is 75 N.
Examples
Understanding forces and motion is crucial in many real-world scenarios. For example, designing elevators requires careful calculation of reaction forces to ensure safety and comfort. Similarly, analyzing pulley systems helps engineers design efficient lifting mechanisms for construction or manufacturing. These principles are also fundamental in understanding vehicle dynamics and the forces involved in transportation.
The reaction force in the lift moving at constant velocity is 2000 N, while with an acceleration of 4.5 m/s², it increases to 2900 N. For the pulley system, the common acceleration is 2.5 m/s² with a tension of 75 N in the string.
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