Calculating Work Done by a Variable Force: Tips and Techniques

[DumSpiroSpero]

Homework Statement

A force F acts over the block as show in the figure. F have a constant magnitude, but has a variable direction. The direction of F, at any given instant, always is directed to the point P.
Evaluate the work done by the force F through the displacement d between the points R and M.
sin 37º = cos53º = 3/5; cos 37º = sin53º = 4/5.

Homework Equations

The Attempt at a Solution

I really can't solve this one. I tried to take only the horizontal axis ( $$F\cos(37º)$$ at R and $$F\cos(53º)$$ at M), because the vertical components are perpendicular to the displacement $$\vec{d}$$. However, since the force have a variable direction, I don't have any idea of how to evaluate the work done between R and M; the angle increases from 37º to 53º, but I don't know how to do that. I never faced any problem in which the force have a variable direction.
I would appreaciate any tips. Thanks!
Answer: $$W = \frac{5}{7}\cdot F \cdot d$$

 

[Simon Bridge]

However, since the force have a variable direction, I don't have any idea of how to evaluate the work done between R and M...

Hint: calculus.

 

[DumSpiroSpero]

$$ W = \int \vec{F} \cdot d\vec{d} = \vec{F} \int d\vec{d} = \vec{F} \cdot \vec{d} = F \cdot d \cdot \cos \theta$$, right? But $$\theta$$ ranges from 37º to 53º, which the definite integral would give sin(53º) - sin (37º) = 4/5 - 3/5 = 1/5 (because the $$\int \cos(x) dx = \sin(x)$$); therefore W = F*d/5. Where I get it wrong?
The work done by a variable force is given by the area under the graph F x d, but the force F is constant, while just the angle are changing...

Also, this is a high school book that explictly says that all questions can be solved without calculus. I'm using the very few knowledge that I have about work and integration that I obtained in a Lang's calculus textbook...
But I would appreciate some explanation even with calculus.

 

[DumSpiroSpero]

https://www.khanacademy.org/math/mu...-find-the-work-done-by-a-vector-field-example
I was watching this video about calculating work using integration. But in this example, F is written in function of t and r also is written in function of t, 0 < t < 2pi, since t changes.
In my problem, the only thing that is changing is the angle and the distance, but I don't know how to relate the distance d to the angle at any instant as a function $$ \vec{d}\theta$$ nor the force respect to the angle $$\vec{F}(\theta)$$ so that I could integrate with the upper bound 53º (in radians) and the lower bound 37º(radians).
Any further help?

 

[Simon Bridge]

You need to know how $\theta$ varies with $x$ all right - this is the lynchpin so just telling you will amount to "doing your homework for you" ;)

Have a play with trigonometry... maybe it is easier to look at x varying with $\theta$?
Try formally writing out how you are defining your coordinate system... and you are not restricted to using the ones implied, so you can try picking different angles and see how the relation for work looks.

 

[DumSpiroSpero]

Thanks! I'm still trying...
Two daft approximations that I have done:
$$F_{m} = \frac{F_{o} + F_{f}}{2} = \frac{F \cdot \cos (37º) + F\cdot \cos (53º)}{2} = F\cdot \frac{7}{10}$$; therefore, $$ W = F_{m} \cdot d = \frac{7}{10} \cdot F \cdot d$$, which is somewhat close to 5Fd/7...
A better approximation:
$$\theta_{m} = \frac{\theta_{o} + \theta_{f}}{2} = \frac{37º+53º}{2} = 45º$$; therefore $$W = F \cdot \cos (\theta_{m}) \cdot d = \frac{\sqrt{2}}{2} \cdot F \cdot d \approx 0,707 \cdot F \cdot d$$, which is closer to 5Fd/7 than the previous approximation

But I'll keep trying the solution with integration until find the 5Fd/7 :)

 

[Simon Bridge]

The last post does not make sense.
If your method does not include calculus then you you need to explain why not: what is your reasoning?
If you do not follow suggestions, nobody can help you.