The force diagram above shows a box accelerating to the right on a horizontal surface of negligible friction. The tension TT is exerted at an angle of 30°30° above the horizontal. If μμ is the coefficient of kinetic friction between the box and the surface, which of the following is a correct mathematical equation derived by applying Newton’s second law to the box?   a=(Tcos(θ)−μMg)/M a=(Tsin(θ)−μMg)/M a=Tcos(θ)−μ(Mg+Tsin(θ))/M   a=Tcos(θ)−μ(Mg−Tsin(θ))/M

please show full solution with formulas and units.

As you will see in a later chapter, forces are vector quantities, and the total force on an object is the vector sum of all forces acting on it. In the figure below, a force ₁ of magnitude 6.80 units acts on a box at the origin in a direction 0 = 29.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the box in the direction of the positive y-axis. Find graphically the magnitude and direction (in degrees counterclockwise from the +x-axis) of the resultant force ₁ + F2. magnitude direction F₂ F₁ i units ° counterclockwise from the +x-axis

In the figures, the masses are hung from an elevator ceiling. Assume the velocity of the elevator is constant. Find the tensions in the ropes (in N) for each case. Note that 8₁ = 35.0°, 0₂ = 55.0°, 03 = 60.0°, m₁ = 3.00 kg, and m₂ = 7.00 kg. (Due to the nature of this problem, do not use rounded intermediate values in your calculations including answers submitted in WebAssign.) (a) (b) T₁ T₂ = LN LM T3 NY M = T₁ = T₂ = = T3 = T₁ M₂ 0₁ T₁ 03 T3 z z N N N T2 N N z z N Ts 0₂ T2 m₁ Ⓡ