Answer:
105 N
Note: This answer is rounded to three significant figures.
Explanation:
To determine the value of the pulling force 'F' required to just start moving a 20 kg block lying on a rough horizontal surface, we need to consider the forces acting on the block, the angle of the applied force, and the coefficient of static friction.
We are given:
- m = 20 kg
- θ = 60°
- μ_s = 0.5
I've attached a free-body diagram for you to view. This was used to answer this question. Let's solve.[tex]\hrulefill[/tex]
To solve, we'll do a sum of forces acting in both the 'x' and 'y' directions. Let's start with the 'x' direction:
[tex]\sum \vec F_x: \vec F \cos(60^\circ)-\vec f_s=0[/tex]
In this scenario, setting the sum of the horizontal forces to zero is appropriate because we're dealing with the point just before the block starts moving. This is a static situation, meaning the block is not yet in motion. Now let's do the 'y' direction:
[tex]\sum \vec F_y: \vec F \sin(60^\circ)+\vec n-\vec w=0[/tex]
In this case, the block is not moving up or down, thus the sum adds to zero. We have the following system of equations:
[tex]\left\{\begin{array}{ccc}\vec F \cos(60^\circ)-\vec f_s=0 & \dots (1)\\\\ \vec F \sin(60^\circ)+\vec n-\vec w=0 & \dots (2)\end{array}\right[/tex]
Plug in the following equations for static friction on weight:
[tex]\boxed{ \begin{array}{ccc} \text{\underline{Formula for Static Friction:}} \\\\ \vec f_s \leq \mu_s \vec n \ \Big(\text{Note: } \vec f_{s_{\text{MAX}}} = \mu_s \vec n\Big) \\\\ \text{Where:} \\ \bullet \ \vecf_s \ \text{is the actual static frictional force} \\ \bullet \ \mu_s \ \text{is the coefficient of static friction} \\ \bullet \ \vec n \ \text{is the normal force} \end{array}}[/tex]
[tex]\boxed{ \left \begin{array}{ccc} \text{\underline{Weight of an Object:}} \\\\ \vec w = mg \\\\ \text{Where:} \\ \bullet \ \vec w \ \text{is the weight of the object (force due to gravity)} \\ \bullet \ m \ \text{is the mass of the object} \\ \bullet \ g \ \text{is the acceleration due to gravity} \end{array} \right.}[/tex]
We have,
[tex]\left\{\begin{array}{ccc}\vec F \cos(60^\circ)-\mu _s \vec n=0 & \dots (1)\\\\ \vec F \sin(60^\circ)+\vec n-mg=0 & \dots (2)\end{array}\right[/tex]
Let's take equation (2) and solve it for 'n':
[tex]\vec F \sin(60^\circ)+\vec n-mg=0\\\\\\\\\therefore \vec n = mg - \vec F \sin(60^\circ) \ \dots (3)[/tex]
Take this new equation and substitute it into equation (1), solve for 'F':
[tex]\Longrightarrow \vec F \cos(60^\circ)-\mu _s\left(mg - \vec F \sin(60^\circ)\right)=0\\\\\\\\\Longrightarrow \vec F \cos(60^\circ)-\mu _smg +\mu _s \vec F \sin(60^\circ)\right)=0\\\\\\\\\Longrightarrow \vec F \cos(60^\circ) +\mu _s \vec F \sin(60^\circ)\right)=\mu _smg[/tex]
[tex]\Longrightarrow \vec F \left(\cos(60^\circ) +\mu _s \sin(60^\circ)\right)=\mu _smg[/tex]
[tex]\therefore \vec F =\dfrac{\mu _smg}{\cos(60^\circ) +\mu _s \sin(60^\circ)}[/tex]
Now plugging in our values, we get:
[tex]\Longrightarrow \vec F =\dfrac{(0.5)(20 \text{ kg})(9.8 \text{ m/s}^2)}{\cos(60^\circ) +(0.5)\sin(60^\circ)}\\\\\\\\\therefore \vec F \approx \boxed{105 \text{ N}}[/tex]
Thus, the pulling force required to just start moving the block is approximately 105 Newtons.
