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The magnitude of the average friction force exerted on the collar (F)= 8.641 N
<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>To calculate the magnitude of the average friction force exerted on the collar we are using the formula,
Here we are given,
k = The spring has a spring constant.
= 25.5 N/m.
= Final length of the spring .
=
= 1.591 m
= The initial length of the spring.
= 1.25−0.60
=0.65 m
y=The collar then travels downward a distance.
= 1.80 m.
m= The mass of the collar.
=3.55 kg
= the velocity of the collar.
= 3.39 m/s.
g = The acceleration due to gravity.
= 9.81 m/s²
We have to calculate the magnitude of the average friction force exerted on the collar = F
Now we put the known values in the above equation, we get;
Or,
Or, F= 8.641 N
From the above calculation we can conclude that,
The magnitude of the average friction force exerted on the collar (F)= 8.641 N
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Question:
The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .
We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:
Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:
Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:
Now, we divide by the distance of the muscle:
Now, we substitute the values:
Now, we solve the operations:
Therefore, the force exerted by the muscles is 23.53 Newtons.
Part B. To determine the force on the pivot we will add the forces we add the vertical forces:
Since there is no vertical movement the sum of vertical forces is zero:
Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:
Now, we plug in the values:
Solving the operations:
Therefore, the force is 73.53 Newtons.