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The mechanism of a stone crusher is along with various dimensions of link in mm. If crank oa rotates at a uniform velocity of 120rpm, determine velocity of point k(jaw) when the crank OA is inclined at an angle of 30 degree to horizontal. What will be the torque required at crank OA to overcome a horizontal force of 40 KN at K?
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The mechanism of a stone crusher is along with various dimensions of l...
Stone Crusher Mechanism

The stone crusher mechanism is a machine designed to reduce large rocks into smaller rocks, gravel, or rock dust. The main components of the mechanism are a crankshaft and a jaw or toggle mechanism. The jaw serves as the mechanism that holds the stones to be crushed and moves them towards the stationary plate. The crankshaft, on the other hand, is responsible for converting the rotary motion of the crank into a reciprocating motion of the jaw.

Dimensions of the Link

To determine the velocity of point K (jaw) when the crank OA is inclined at an angle of 30 degrees to the horizontal, we need to consider the dimensions of the link. Let's assume that the length of the crank OA is 'a' mm, the length of the link AB is 'b' mm, and the length of the link BC (jaw) is 'c' mm.

Uniform Velocity of Crank OA

Given that the crank OA rotates at a uniform velocity of 120 rpm, we can calculate the angular velocity of the crank. Angular velocity is given by the formula ω = 2πn/60, where 'n' is the speed in rpm. Therefore, the angular velocity ω of the crank OA is:

ω = (2π * 120)/60 = 4π rad/min

Velocity of Point K

To determine the velocity of point K when the crank OA is inclined at an angle of 30 degrees to the horizontal, we can use the concept of relative velocity. The velocity of point K can be calculated by considering the velocities of points O and A.

Let v_OA be the velocity of point A relative to point O. The magnitude of v_OA is given by v_OA = a * ω, where 'a' is the length of the crank OA. Since the crank OA is inclined at an angle of 30 degrees to the horizontal, the vertical component of v_OA can be calculated as v_OA * sin(30).

Similarly, let v_KA be the velocity of point A relative to point K. The magnitude of v_KA is given by v_KA = c * ω, where 'c' is the length of the link BC (jaw). The vertical component of v_KA can be calculated as v_KA * sin(30).

The velocity of point K can be obtained by subtracting the vertical component of v_OA from the vertical component of v_KA.

Torque Required to Overcome Horizontal Force at K

To calculate the torque required at the crank OA to overcome a horizontal force of 40 kN at point K, we can use the equation τ = F * r, where τ is the torque, F is the force, and r is the distance between the line of action of the force and the axis of rotation.

In this case, the distance 'r' is equal to the length of the crank OA, which is 'a'. Therefore, the torque τ required is given by τ = 40 kN * a.

Explanation

The stone crusher mechanism consists of a crankshaft and a jaw or toggle mechanism. The crankshaft converts the rotary motion of the crank into a reciprocating motion of the jaw, which moves the stones towards the stationary plate for crushing.

To determine the velocity of the jaw when
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The mechanism of a stone crusher is along with various dimensions of link in mm. If crank oa rotates at a uniform velocity of 120rpm, determine velocity of point k(jaw) when the crank OA is inclined at an angle of 30 degree to horizontal. What will be the torque required at crank OA to overcome a horizontal force of 40 KN at K?
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The mechanism of a stone crusher is along with various dimensions of link in mm. If crank oa rotates at a uniform velocity of 120rpm, determine velocity of point k(jaw) when the crank OA is inclined at an angle of 30 degree to horizontal. What will be the torque required at crank OA to overcome a horizontal force of 40 KN at K? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The mechanism of a stone crusher is along with various dimensions of link in mm. If crank oa rotates at a uniform velocity of 120rpm, determine velocity of point k(jaw) when the crank OA is inclined at an angle of 30 degree to horizontal. What will be the torque required at crank OA to overcome a horizontal force of 40 KN at K? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mechanism of a stone crusher is along with various dimensions of link in mm. If crank oa rotates at a uniform velocity of 120rpm, determine velocity of point k(jaw) when the crank OA is inclined at an angle of 30 degree to horizontal. What will be the torque required at crank OA to overcome a horizontal force of 40 KN at K?.
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