The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other end is connected to box 3. The weights of the three boxes are W1 = 51.0 N, W2 = 36.6 N, and W3 = 29.8 N. Determine the magnitude of the normal force that the table exerts on box 1.
A 260-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.870, and the log has an acceleration of 0.800 m/s2. Find the tension in the rope.
To determine the magnitude of the normal force that the table exerts on box 1, you need to consider the forces acting on box 1.
First, let's consider the weight of box 1. The weight of an object is the force acting on it due to gravity, and it is given by the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.
In this case, W1 = 51.0 N. We can use the formula W = m * g to find the mass of box 1. Rearranging the formula, we have m = W / g.
The acceleration due to gravity, g, is approximately 9.8 m/s². So we have m = 51.0 N / 9.8 m/s² ≈ 5.20 kg.
Now, let's consider the forces acting on box 1. There are two forces: the weight of box 1, acting downwards, and the normal force exerted by the table, acting upwards.
Since box 1 is in equilibrium (not accelerating), the sum of the forces acting on it must be zero.
The weight of box 1, acting downwards, is 51.0 N.
The normal force exerted by the table, acting upwards, cancels out the weight of box 1. So the normal force is also 51.0 N.
Therefore, the magnitude of the normal force that the table exerts on box 1 is 51.0 N.
To determine the magnitude of the normal force that the table exerts on box 1, we need to consider the forces acting on box 1.
1. First, let's consider the weight of box 1, which acts vertically downward. The weight is given as W1 = 51.0 N.
2. Since box 1 is resting on the table, there is a normal force exerted by the table on box 1. The normal force acts perpendicular to the table surface and opposes the weight of box 1.
3. Since the boxes are at rest, the net force acting on each box must be zero.
4. We can start by considering box 2. The force of tension in the rope acts upwards on box 2 and is equal to the weight of box 3, i.e., T = W3. Also, the weight of box 2 acts vertically downward.
5. Since the rope is massless and frictionless, the tension in the rope is the same throughout. Hence, the force of tension is also acting downwards on box 1.
6. Considering the forces acting on box 1, we have the weight of box 1 acting downward and the tension in the rope acting downward. The normal force exerted by the table on box 1 acts upward.
7. Now, we can set up the equation to represent the forces acting on box 1:
Net downward force = Weight of box 1 + Tension in rope
Net downward force = W1 + T
Since the net force acting on box 1 is zero, we have:
Weight of box 1 + Tension in rope = 0
Substituting the known values, we get:
51.0 N + T = 0
Solving for T, we find:
T = -51.0 N
Note that the negative sign indicates that the tension is acting downward.
8. Now, considering the vertical forces on box 1, we have:
Net upward force = Normal force - Tension in rope
Net upward force = N - T
Also, since the net force acting on box 1 is zero, we have:
Normal force - Tension in rope = 0
Substituting the known values, we get:
N - (-51.0 N) = 0
Solving for N, we find:
N = -(-51.0 N)
Simplifying, we get:
N = 51.0 N
Hence, the magnitude of the normal force that the table exerts on box 1 is 51.0 N.