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franstirlacci
13.02.2022 •
Physics
Calculate the force per unit mass on the earth Surface (Take G=6.67X10Nm ² kg^-2) mass of earth=5.98 x 10^24 kg, radius of earth=6.38 x 10^6m
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Ответ:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
∑Fᵧ = maᵧ T₁ - m₂g = m₂aᵧNote that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
T₁ = m₂aᵧ + m₂gT₁ = m₂(a + g)We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
∑Fₓ = maₓ F_f - F_g sinΘ = maₓThe normal force is equal to the x-component of the force of gravity.
(F_n · μ_k) - m₁g sinΘ = m₁aₓ (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ [2.539595871] - [-58.0962595] = 6aₓ 60.63585537 = 6aₓ aₓ = 10.1059759 m/s²Now let's go back to this equation:
T₁ = m₂(a + g)We have 3 known variables and we can solve for the tension force.
T = 2(10.1059759 + 9.8)T = 2(19.9059759)T = 39.8119518 NThe tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.