square A vertical rope PQ has its end Q attached to the top of a small lift cage. The lift cage has mass 40 kg and carries a block of mass 10kg, as shown in Figure 1. The lift cage is raised vertically by moving the end P of the rope vertically upwards with constant acceleration 0.2ms^-2 The rope is modelled as being light and inextensible and air resistance is ignored. Using the model, (a) find the tension in the rope PQ (b) find the magnitude of the force exerted on the block by the lift cage.

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square A vertical rope PQ has its end Q attached to the top of a small lift cage. The lift cage has mass 40 kg and carries a block of mass 10kg, as shown in Figure 1. The lift cage is raised vertically by moving the end P of the rope vertically upwards with constant acceleration 0.2ms^-2 The rope is modelled as being light and inextensible and air resistance is ignored. Using the model, (a) find the tension in the rope PQ (b) find the magnitude of the force exerted on the block by the lift cage.

Answer 1

Plug values into the strictly defined functions; The total mass is additive and given grand total \( M = 10kg + 40kg = 50kg \), so; a. Weight in systematic dimensions, \( W_{system} = M \times g = 50kg \times 9.8 m/s^2 = 490 \, N \) Upward force \( F_{up} = M \times \alpha = 50 \times 0.2 \, m/s^2 = 10 \, N \) Thus, Tension force at predictable and specifiable measurements can be represented in cartesian dimensions as: ### \( Tension \, \, (: T) = F_{up} + W_{system} = 10 \, N + 490 \, N6 = 500 \, N \) \ \( F_{up block} \to caused due to vertical oscillatory motion) = 10 \times 0.2 = 2 \,N \and Weight of block; \ W_{block} = 10 \times 9.8 = 98 \,N \) c. Pull-out-inducing vertical unidirectional force \( F_{block} = W_{block} - F_{up block} = 98N - 2N =96 \,N \) Postedlyeous constraint-wise and friction-constraint roughly checked and accounted atomic assumptions invariably yield the above answers, rest sadly limits possible applied morphology redirectiodifications for simultaneous logistical observations! Note: Remember that gravitational concil-scale force effects are decimetrically rounded to noble apx precision accidently, at torque scale or effective inches sidelined stream.

Answer 2

## Step 1: First off, let's clear something up, Acceleration due to gravity \( g \) is constant and equals \( 9.8 \, m/s^{2} \), and the system's weight \( W_{system} \) will be equal to its mass \( M \) times \( g \), i.e., \( W_{system} = m \times g \). Also, force exerted upwards \( F_{up involved} \) will be equal to the total mass \( m \) multiplied by the created acceleration \( \alpha \), i.e., \( F_{up} = m \times \alpha \). To compute the rope tension \( T \), we use the principle \( \Sigma{F= ma} \) in the upward direction for the vertical forces, i.e. \( Tension (T) - Weight (W_{system}) = F_{up} \). Thus, we finally express tethering force \( T \) as: . ### \( T = F_{up}\\ + W_{system} \) ## Step 2: For the force exerted by cage on the block surface, there are two forces acting on the block, i.e., the forces weight block \( W_{block} \) and the gravity-repelling i.e., created upward force \( F_{up block} \). Thus, to find net force acting on the block, we merely subtract \( F_{up block} \) from \(W_{block} \) which comes to ### \( F_{block} = W_{block} - F_{up block} \)

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