Change in acceleration with change of angle of applied force

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The discussion centers on how the acceleration of a block changes with an increase in the angle of the applied force. With a coefficient of kinetic friction (μ_k) set at 0.5 and an applied force of 10 Newtons, the mathematical analysis indicates that increasing the angle by 5 degrees leads to an increase in acceleration until a critical angle of approximately 28.955 degrees is reached, after which acceleration begins to decrease. The memo's assertion that acceleration would decrease is challenged, as the calculations and graph suggest otherwise. The conclusion drawn is that for the given conditions, the acceleration increases with the angle until it surpasses 48.76 degrees. The discussion confirms that the original analysis is correct, aligning with the mathematical findings.
GreenOlive
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Homework Statement
How will the acceleration of the block change if the angle of the applied force is increase by 5 degrees? Write Increase, Decrease or Stay the same.
Relevant Equations
Newton 2, Fnet = ma
friction = N * μ
Untitled Diagram.png

Let ##μ_k## = 0.5
##F_a## = 10 Newtons
##\theta## is the angel of the Applied force.
How will the acceleration of the block change if the angle of the applied force is increase by ##5^o##? Write Increase, Decrease or Stay the same.

Recently we were discussing a question similar to this in class, I believe the acceleration should INCREASE however the memorandum for the question said it would DECREASE. I would like to just confirm my maths is correct (This is not the first time I have seen such a question).

This is my general solution.Let right be positive.

$$F_{net} = ma$$
$$F_{net} = F_{a,x} - f$$
$$ma = F_{a,x} - (N \times μ_k)$$
$$ma = F_a \times cos\theta - ((F_g-F_{a,y}) \times μ_k) $$
$$ma = F_a \times cos\theta - ((F_g-(F_a\times sin\theta)) \times μ_k) $$

As ##\theta## is the only variable changing let's isolate the expressions it is part of.

$$ma = F_a \times cos\theta + F_a\times sin\theta \times μ_k - F_g \times μ_k $$
$$ma = F_a (cos\theta + sin\theta \times μ_k) - F_g \times μ_k $$

lets graph ##(cos\theta + sin\theta \times μ_k) ##
desmos-graph.png

Desmos graph

Looking at the graph we can see ##(cos\theta + sin\theta \times μ_k)##increases from 0 until its maximum at 28.955 degrees from which it starts to decrease.
And as $$a = \frac{F_a (cos\theta + sin\theta \times μ_k) - F_g \times μ_k}m $$

Therefore by increasing the angle of the applied force by 5 degrees the acceleration will INCREASE.
By looking closer at the graph you could see that ##\theta## would need to be greater than 48.76 degrees for the acceleration to Descrease(or less than 5).

Who is Incorrect, the Memo or I?
 
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Your graph, if proportional to acceleration, still shows positive acceleration when θ is > 90 degrees, which seems to imply that the block will move forward when you're applying negative force to it. That intuitively suggests something amiss.
It being positive for a while with a negative θ is OK. A force pulling a bit down can still move the block until the added friction eventually prevents any motion. So the left side of the graph seems ok.
 
Can we create a limit: -90 < ##\theta## < 90. Angels not in this range would imply a negative applied force and then positive friction(if right is still positive) so you would have to change the equation.
 
The answer depends on the value of ##\mu##. For ##\mu = 0.5##, your answer is correct.
 
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@TSny I did include that under the image. Thanks for the confirmation
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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