Respuesta :
Answer:
d) 82 N
Explanation:
Draw a free body diagram of the block. There are 5 forces:
Weight mg pulling down,
Normal force N pushing to the right,
Friction force Nμ pushing down,
Applied force F pushing up,
and applied force P pushing left.
Sum of forces in the x direction:
∑F = ma
N − P = 0
N = P
Sum of forces in the y direction:
∑F = ma
F − mg − Nμ = 0
F = mg + Nμ
Substitute and solve:
F = mg + Pμ
F = (5 kg) (10 m/s²) + (80 N) (0.4)
F = 82 N
Answer:
d) 82 N
Note: Using a value of 9.8 m/s² yields an answer of 81 N. However, if we use 10 m/s² for the acceleration due to gravity, the result aligns with option (d), giving a value of 82 N.
Explanation:
To determine the maximum value of the upward force 'F_V' that can be applied without causing the block to move, we need to consider the static frictional force between the block and the wall, which is the key factor preventing the block from sliding down due to its own weight.
We are given:
- m = 5 kg
- F_H = 80 N
- μ_s = 0.4
I've attached a free-body diagram for you to view. This was used in answering this question, please view. Now, let's solve. [tex]\hrulefill[/tex]
To solve this question we will do a sum of forces. We need to consider the forces acting in both the 'x' and 'y' directions. Let's start with the 'x' direction:
[tex]\sum \vec F_x: \vec F_H-\vec n = 0[/tex]
Now the 'y' direction:
[tex]\sum \vec F_y: \vec F_{V_{\text{MAX}}}-\vec f _s - \vec w = 0[/tex]
Thus, we have the following system of equations:
[tex]\left\{\begin{array}{ccc}\vec F_H-\vec n = 0 & \dots (1)\\\\\vec F_{V_{\text{MAX}}}-\vec f _s - \vec w = 0 & \dots (2) \end{array}\right[/tex]
Using the following equations for static friction and weight:
[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]
[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]
We have,
[tex]\left\{\begin{array}{ccc}\vec F_H-\vec n = 0 & \dots (1)\\\\\vec F_{V_{\text{MAX}}}-\mu _s \vec n - mg = 0 & \dots (2) \end{array}\right[/tex]
From equation (1) we find that the normal force is equal to the applied horizontal force of 80 Newtons. Knowing this fact, let's solve for 'F_V' in equation (2):
[tex]\Longrightarrow \vec F_{V_{\text{MAX}}}-\mu _s \vec n - mg = 0\\\\\\\\\Longrightarrow \vec F_{V_{\text{MAX}}} = \mu _s \vec n + mg[/tex]
Plug in our given values:
[tex]\Longrightarrow \vec F_{V_{\text{MAX}}} = (0.4)(80 \text{ N}) + (5 \text{ kg})(9.8 \text{ m/s}^2)\\\\\\\\\Longrightarrow \vec F_{V_{\text{MAX}}} = 32 \text{ N} + 49 \text{ N}[/tex]
[tex]\therefore \vec F_{V_{\text{MAX}}} = \boxed{81 \text{ N}}[/tex]
Thus, the maximum value of the upward force 'F_V' that can be applied without causing the block to move upward is 81 Newtons. This does not align with any of the options given. Using a less accurate and rounded value of the acceleration due to gravity yeilds a result of 82 Newtons.
[tex]\Longrightarrow \vec F_{V_{\text{MAX}}} = (0.4)(80 \text{ N}) + (5 \text{ kg})(10 \text{ m/s}^2)\\\\\\\\\Longrightarrow \vec F_{V_{\text{MAX}}} = 32 \text{ N} + 50 \text{ N}[/tex]
[tex]\therefore \vec F_{V_{\text{MAX}}} = \boxed{82 \text{ N}}[/tex]
Under this assumption, the correct answer is option (d).