Solving Tension Force: Angle & Magnitude

  • Thread starter Thread starter RedBurns
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
A ball hanging from a rearview mirror experiences tension in the string due to the car's acceleration. The car accelerates at 1.0 m/s², and the ball's mass is 4 kg, leading to equations for horizontal and vertical forces. The weight of the ball is calculated as 39.2 N, and the total force considering acceleration is 43.2 N. To find the tension, the equations T sin(theta) = m a and T cos(theta) = mg can be used, with the suggestion to divide one by the other to simplify. Visualizing the forces as a triangle can help determine the angle theta.
RedBurns
Messages
24
Reaction score
2
A ball hangs from a string on the rear view mirror of a car. The car is accelerating, causing the string to be at an angle theta with the vertical. If the car is accelerating at 1.0m/s2 and the mass of the ball is 4kg, what is the magnitude of the tension on the string, and at what angle is it?

Fx) -T* Sin (theta) = MA
Fy) T*Cos(theta) -w=0

MA= 4 * 9.8 = 39.2
W= 39.2 + (4 * 1.0)= 43.2
--------------------------------
Fx) -T * Sin (theta)= 39.2

Fy) T * Cos (Theta) -43.2 = 0
Fy) T * Cos (Theta) = 43.2

Could somone point me in the right direction for the next step in finding T? I'm not very strong in math and I'm not sure were to go from here.
 
Physics news on Phys.org
RedBurns said:
Fx) -T* Sin (theta) = MA
Fy) T*Cos(theta) -w=0
At this point, you're doing fine. I'm not sure what you were doing after that.

Rewrite your equations like this:
T \sin\theta = m a
T \cos\theta = w = mg

To solve these two equations, try dividing one by the other. See what happens.
 
acceleration a = 1.0 m/s/s ...
This "a" has ALMOST NOTHING to do with "g" = 9.8 N/kg !

the weight = Force by gravity = mg , = T cos(theta) .
T sin(theta) = m a .

Try drawing a triangle of these components, and see what tan(theta) is ...
 
Thanks! Thats what I needed.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top