Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
1 answer:
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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
The tension in the rope is 229.37 N.
Explanation:
Given:
Mass of the block is,
Coefficient of static friction is,
Angle of inclination is, °
Draw a free body diagram of the block.
From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.
Forces acting are and normal
. Now, there is no motion in the direction perpendicular to the inclined plane. So,
Consider the direction along the inclined plane.
The forces acting along the plane are and frictional force,
, down the plane and tension,
, up the plane.
Now, as the block is at rest, so net force along the plane is also zero.
Therefore, the tension in the rope is 229.37 N.
