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-BARSIC- [3]
3 years ago
14

In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pu

shing at an angle of 30 degrees a distance L from the axis. 1)In which case is the torque due to the force about the rotation axis biggest?
Physics
1 answer:
pashok25 [27]3 years ago
7 0

Answer:

The answer is c) same.

Explanation:

When a force is applied at some point on a rigid body, the body makes a rotational movement about the axis. Said property of the force applied to rotate the body is calculated with a magnitude known as torque, which is defined as the moment of force at a capacity of that force to make a rotation around a specific point.

for case 1, the magnitude of the torque is equal to:

τ = F*(L/2)

For case 2, the torque magnitude equals:

τ = F*sin(30°) = F*(L/2)

It can be seen that in both cases it is the same torque.

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Complete Question

The complete question is shown on the first uploaded image

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Looking at  the diagram uploaded we see that there are two  forces acting along the y-axis on the fixed support  

   These force are  500 N  and  A_{(y)} [ i.e the force acting along the same direction with 500 N   ]

Hence the sum of forces along the x axis is mathematically represented as

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Looking at  the diagram uploaded we see that there are two  forces acting along the z-axis on the fixed support  

       These force are  600 N  and  A_{(z)} [ i.e the reactive force of  600 N  ]

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        A_{(z)}  - 600  = 0

=>     A_{(z)}  = 600  \ N

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    \sum M_x  = M_x  - 500 (0.75 + 0.5) + 600 ( 1 ) = 0

=>   M_x  = 1225 \  N\cdot m

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    \sum M_y  = M_y  - 400 (0.75 ) + 600 ( 0.75 ) = 0

=>   M_y  = 750 \  N\cdot m

Generally taking moment about A along the z-axis we have that

    \sum M_z  = M_z = 0

=>   M_z  = 0 \  N\cdot m

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