Force exerted on a pivot point

  • Thread starter Thread starter benshields
  • Start date Start date
  • Tags Tags
    Force Pivot Point
AI Thread Summary
To determine the force exerted on a 4.0 m horizontal plank by the second pivot point, the problem involves analyzing the plank's equilibrium. The total weight of the plank, which is 24.0 kg, must be balanced by the forces at the two pivot points. By taking moments around a convenient point, it can be established that the torques from each pivot are equal and opposite. The center of mass is assumed to be at the midpoint of the plank, 2.0 m from either end. This analysis leads to the conclusion that the forces at the pivots must sum to equal the weight of the plank, ensuring both translational and rotational equilibrium.
benshields
Messages
3
Reaction score
0

Homework Statement


A horizontal plank 4.0 m long and having mass 24.0 kg rests on two pivot points, one left at the end and a second one 1.0 m from the right end. What is the magnitude of the force exerted on the plank by the second pivot point?

Homework Equations

The Attempt at a Solution


I'm not even sure how to attempt this problem
 
Physics news on Phys.org
Draw a large diagram and mark on the forces and their location.

Take moments about a convenient point.
 
Draw a free body diagram.
Assuming the mass distribution is same, take the centre of mass at the centre i.e at 2.0 metre.
The plank is in translational and rotational equilbrium.
Thus, the sum of magnitude of the forces at the pivot is equal to the wieght of the plank.
And, the moments, or the torque exerted by each of the pivot is equal and opposite thus their magnitudes are equal.
Take the origin as the centre of mass.

PS:- Torque= Force*distance of action from the origin (when force is perpendicular)
In vector notation torque is the cross product of radius and force. But the above equation is enough here

Hope it helps :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top