Pls help on this one
2 answers:
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Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0
Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
Answer:
The friction force is 250 N
Explanation:
The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):
In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:
(1)
We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:
- the pull, forward, F = 250 N
- the friction force, backward,
Given (1), we have
So the force of friction must be equal to the pull: