A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initi
ally, the fluid level is such that the block is at a depth d and the tension in the thread is T. Then, the fluid level is decreased such that the depth is 0.5d. What is the tension in the thread when the block is at the new depth?
1 answer:
Answer:
(C) T
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence the tension.
Note: The figure of question is added in the attachment
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Answer:
a)
b)
Explanation:
In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).
After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:
We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:
When solving for the normal force we get:
we know that
W=mg
and
so after substituting we get that
N=F sin θ +mg
We also know that the kinetic friction is defined to be:
so we can find the kinetic friction by substituting for N, so we get:
Now we can find the sum of forces in x:
so after analyzing the diagram we can build our sum of forces to be:
we know that:
so we can substitute the equations we already have in the sum of forces on x so we get:
so now we can solve for the force, we start by distributing so we get:
we add to both sides so we get:
Nos we factor F so we get:
and now we divide both sides of the equation into so we get:
which is our answer to part a.
Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:
and
F cos θ = f
when substituting one into the other we get:
which can be solved for the static friction coefficient so we get:
which is the answer to part b.
