Part A
1 answer:
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The force on the fry is
Explanation:
We can find the force acting on the fry by using Newton's second law:
where
F is the net force on the fry
m is its mass
a is its acceleration
For the fry in this problem,
Therefore, the force exerted on the fry is
Learn more about Newton's second law here:
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Answer:
F = 24 N
Explanation:
In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end
Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity
∑ τ = 0
F l -x W = 0
F =
let's calculate
F = 4/100 600
F = 24 N
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.