Question

Two horizontal forces F1 and F2 act on a 4 kg disk that slides over frictionless ice, on which an xy coordinate system is laid out. F1 is in the positive direction of x axis and has magnitude = 7 N. F2 has magnitude 9 N. The image below gives velocity as a function of time along x axis. What is angle between F1 and F2?

Answer 1

Answer:

$56.3^{\circ}$

Explanation:

The figure is missing, find it in attachment.

The figure shows the velocity along the x-axis, $v_x$, versus time. The slope of the line gives us the acceleration along the x-axis, which is therefore given by:

$a_x = \frac{\Delta v_x}{\Delta t}=\frac{5-2}{3-2}=3 m/s^2$

Now we can apply Newton's second law along the x-axis:

$F_x = m a_x$

where

$F_x$ is the resultant force along the x-axis

m = 4 kg is the mass of the disk

$a_x = 3 m/s^2$ is the acceleration along this axis

Solving,

$F_x = (4)(3)=12 N$

So now we know that this is the component of resultant force along the x-axis. The resultant force is given by two forces:

F1, which acts along the x-direction

F2, which acts at a certain angle $\theta$

So the component of the resultant along the x-axis can be written as

$F_x = F_1 + F_2 cos \theta$

where we know:

$F_1 = 7 N\\F_2 = 9 N$

Solving for $\theta$, we find:

$cos \theta = \frac{F_x-F_1}{F_2}=\frac{12-7}{9}=0.555\\\theta =cos^{-1}(0.555)=56.3^{\circ}$