In the system shown in the figure below, a horizontal force Fvecx acts on the 7.00 kg object (m1). The horizontal surface is frictionless.
(a) For what values of Fx does the 1.00 kg object (m2) accelerate upward?
To determine the values of Fx that cause the 1.00 kg object (m2) to accelerate upward, we need to consider the forces acting on both objects.
Let's analyze the forces acting on the objects:
For the 7.00 kg object (m1):
- There is a vertical force due to gravity acting downwards, equal to the weight, which can be calculated as Fg1 = m1 * g, where m1 = 7.00 kg and g = 9.8 m/s^2.
- Since the horizontal surface is frictionless, there are no other horizontal forces acting on m1.
For the 1.00 kg object (m2):
- There is a vertical force due to gravity acting downwards, equal to the weight, which can be calculated as Fg2 = m2 * g, where m2 = 1.00 kg and g = 9.8 m/s^2.
- There is a tension force T acting upwards, which is transmitted through the pulley and affects m1 and m2 equally.
The tension force T can be determined using the relationship between the acceleration and the net force on m2:
T - Fg2 = m2 * a
Since we want the 1.00 kg object (m2) to accelerate upward, the net force acting on it must be upwards. This means that T must be greater than Fg2.
Similarly, the net force acting on the 7.00 kg object (m1) can be calculated using its acceleration, which is equal in magnitude but opposite in direction to the acceleration of m2:
Fg1 - T = m1 * (-a)
Since the horizontal surface is frictionless, the only horizontal force acting on m1 is Fx. Therefore, we can write:
Fx = m1 * (-a)
From these equations, we can see that the magnitude of Fx must be greater than Fg1 for the 7.00 kg object (m1) to accelerate in the direction of Fx.
To summarize:
For the 1.00 kg object (m2) to accelerate upward, we need Fx > Fg2.
For the 7.00 kg object (m1) to accelerate in the direction of Fx, we need Fx > Fg1.
Plug in the values and calculate Fg1 and Fg2 to determine the specific values of Fx that meet these conditions.