- #1
Rijad Hadzic
- 321
- 20
Homework Statement
Two boxes with masses [itex] m_1 = 4 kg [/itex] and [itex] m_2 = 10 kg [/itex] are attached by a massless cord passing over a frictionless pulley as shown in in my picture. The incline is frictionless and thetha = 30 degrees
Homework Equations
f = ma
The Attempt at a Solution
[/B]
Tension 1 is above mass 1 and tension two is to the right of mass 2
So I find [itex] F_gm1 = 9.8(4 kg) = 39.2 N [/itex] that means [itex] t_1 = 39.2 N [/itex]
Now on mass 2 I have 3 forces acting on it. Normal force perpendicular to the surface, force of gravity going straight down, and tension force connecting the rope and box.
[itex] F_g = (9.8)(10 kg) = 98 N [itex] and this force has components:[itex] F_g\cos\Theta [/itex] perpendicular to surface of contact, = to my normal force, and [itex] F_g\sin\Theta [/itex] parallel to my surface. This force gives me the tension in the rope.
Is what I am saying here correct? It makes sense to me since [itex] F_g\sin\Theta[/itex] is parallel in the opposite direction of tension, and [itex] F_g\cos\Theta [/itex] is parralel opposite direction normal force.
So I calculate [itex] F_g\cos\Theta = 84.87048957 N, F_g\sin\Theta = 49 N [/itex]
These components make sense because squaring both, then adding, then taking the square root gives me my [itex] F_g = 98 N [/itex]
So that gives me [itex] F_n = 84.87048957 N [/itex]
Tension = [itex] F_t = 49 N [/itex]
So I think I've dealt with all the forces here. The next part of my question asks:
What is the magnitude of the acceleration of the boxer?
So I know f = ma, and the box on the right has a tension of 49 N, and the box on the left has a tension of 39.2 N. So wouldn't my Net forces be 49 N - 39.2 N, going in the same direction as [itex] F_g\sin\Theta [/itex] ? Since that has a larger magnitude (49) opposed to (39.2) the force of gravity on box 1?
So net force = 9.8 N going in the direction of [itex] F_g\sin\Theta[/itex], now I can use f=ma
So my question here is, is the m in that equation = to both of the masses that are being involved??
so would it be 9.8 = 14a, 9.8/14 = a
would that be the magnitude of acceleration?