a person is pulling a 100 lb box across a smooth horizontal surface. what is the acceleration of the box if they are pulling upward at an angle of 20° with a force of 75 lbs
Well, I know that there is no friction involved because it is a smooth surface and that n+TsinΘ-mg = 0
but I'm not sure why TcosΘ = m * asubX (means a with a lower exponent of x [or subscript])
I've been having trouble this whole semester but I'm starting to understand that I need to be clever and find values of variables in order to solve other variables, or substitute/manipulate one equation/solve for a variable algebraically in order to find another value of an equation. (I am horrible in algebra btw).
To understand why TcosΘ = m * asubX in this scenario, let's break down the forces acting on the box.
In the given problem, the person is pulling the box at an angle of 20° above the horizontal surface with a force of 75 lbs. This force can be separated into two components: one perpendicular to the surface (TcosΘ) and one parallel to the surface (TsinΘ).
The force that opposes the box's movement is the force of gravity, which is equal to the weight of the box, 100 lbs in this case. We can represent this force as mg, where m is the mass of the box (which we'll assume to be constant) and g is the acceleration due to gravity (approximately 32.2 ft/s²).
Now, let's consider the horizontal motion of the box. Since there is no friction, the only force that causes horizontal acceleration is the component of the applied force parallel to the surface, TsinΘ. Therefore, the equation for the horizontal motion of the box is:
TsinΘ = m * asubX
In this equation, TsinΘ represents the horizontal component of the pulling force, m represents the mass of the box, and asubX represents the acceleration in the horizontal direction.
On the other hand, the vertical motion of the box is influenced by the vertical component of the applied force (TcosΘ) and the force of gravity (mg). Since the box is on a smooth surface, the vertical acceleration is zero (asubY = 0). Using Newton's second law in the vertical direction, we have:
TcosΘ - mg = 0
In this equation, TcosΘ represents the vertical component of the pulling force, and mg represents the force of gravity.
By solving these equations simultaneously, we can find the values of TcosΘ and asubX. Note that while TcosΘ is equal to mg, asubX depends on the specific values of TsinΘ and m.