A 3.2 Kg box is pressed up against a wall by a person’s arm that makes a 35-degree angle with the wall. The box slides down the wall with a uniform speed. There is a coefficient of friction of 0.45 between the box and the wall.
A. What is the magnitude the force that the person is using to push on the box?
B. If the box is sliding up the wall at a uniform speed, what is the magnitude of the force the person is using to push on the box.
To find the magnitude of the force that the person is using to push on the box, we need to consider the forces acting on the box.
A. The forces acting on the box are:
1. Weight (mg): The weight of the box, given by the mass (m) multiplied by the acceleration due to gravity (g). In this case, the weight is calculated as W = 3.2 kg × 9.8 m/s^2 = 31.36 N.
2. Normal force (N): The force exerted on the box by the wall in the perpendicular direction, which is equal and opposite to the weight of the box (mg) in this case. So, the normal force is N = 31.36 N.
3. Frictional force (f): The force opposing the motion of the box along the wall, given by the product of the coefficient of friction (μ) and the normal force (N). So, the frictional force is f = μN = 0.45 × 31.36 N = 14.112 N.
4. Force applied by the person (F_p): This force is applied at an angle of 35 degrees with respect to the wall.
To find the magnitude of the force applied by the person (F_p), we can use the equilibrium condition since the box is sliding down the wall at a uniform speed. In this case, the vertical component of the force applied by the person must balance the weight of the box, and the horizontal component of the force applied by the person must balance the frictional force.
The vertical component of the force applied by the person is given by F_p * sin(35 degrees), and it should balance the weight (mg) of the box:
F_p * sin(35 degrees) = mg
F_p = mg / sin(35 degrees) = 31.36 N / sin(35 degrees) ≈ 57.55 N
Therefore, the magnitude of the force that the person is using to push on the box is approximately 57.55 N.
B. If the box is sliding up the wall at a uniform speed, the forces acting on the box will remain the same. However, the direction of forces will be opposite compared to the previous case.
To find the magnitude of the force applied by the person (F_p) when the box is sliding up the wall at a uniform speed, we can again use the equilibrium condition:
The vertical component of the force applied by the person should now balance the weight (mg) of the box:
F_p * sin(35 degrees) = mg
F_p = mg / sin(35 degrees) = 31.36 N / sin(35 degrees) ≈ 57.55 N
Therefore, the magnitude of the force that the person is using to push on the box, even when the box is sliding up the wall at a uniform speed, is approximately 57.55 N.
To find the magnitude of the force that the person is using to push on the box, we can break it down into two components: the vertical component and the horizontal component.
A.
First, let's calculate the vertical component of the force which counteracts the weight of the box. We know that the weight of the box is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, the weight of the box is W = 3.2 kg * 9.8 m/s^2 = 31.36 N.
Since the box is sliding down the wall with a uniform speed and there is no vertical acceleration, the vertical component of the force (Fv) must be equal in magnitude and opposite in direction to the weight of the box.
Therefore, Fv = 31.36 N.
Next, let's find the horizontal component of the force using the coefficient of friction. Since the box is sliding down the wall with a uniform speed, the net force in the horizontal direction must be zero. The horizontal component of the force (Fh) is given by the equation Fh = μ * Fv, where μ is the coefficient of friction.
Using the given coefficient of friction μ = 0.45 and the vertical component of the force calculated previously (Fv = 31.36 N), we can determine the horizontal component of the force.
Fh = 0.45 * 31.36 N ≈ 14.112 N.
To find the magnitude of the force that the person is using to push on the box, we can use the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Therefore, the magnitude of the force (F) is given by the equation F = √(Fh^2 + Fv^2).
Plugging in the values we calculated, we get:
F = √(14.112 N^2 + 31.36 N^2) ≈ √(197.836 N^2) ≈ 14.07 N.
So, the magnitude of the force that the person is using to push on the box is approximately 14.07 N.
B.
If the box is sliding up the wall at a uniform speed, the only change in the problem is the direction of the frictional force. The rest of the calculation remains the same.
Since the box is sliding up the wall with a uniform speed, the vertical component of the force (Fv) must be equal in magnitude but opposite in direction to the weight of the box, which is 31.36 N.
The horizontal component of the force is still given by the equation Fh = μ * Fv, where μ is the coefficient of friction (0.45) and Fv is the vertical component of the force (31.36 N).
Plugging in the values, we get:
Fh = 0.45 * 31.36 N ≈ 14.112 N.
To find the magnitude of the force that the person is using to push on the box, we can again use the Pythagorean theorem:
F = √(Fh^2 + Fv^2) = √((14.112 N)^2 + (31.36 N)^2) ≈ √(197.836 N^2) ≈ 14.07 N.
So, the magnitude of the force that the person is using to push on the box when it is sliding up the wall at a uniform speed is approximately 14.07 N, which is the same as in the previous case.