A 50-kg crate is pulled across the ice with a rope. A force of 100 N is applied at an angle of 38° with the horizontal. Neglecting friction, calculate the acceleration of the crate AND the upward force the ice exerts on the crate as it is pulled.
To find the acceleration of the crate, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Step 1: Convert the force and angle to horizontal and vertical components.
The horizontal component, F_x, can be found by multiplying the force (100 N) by the cosine of the angle (38°):
F_x = 100 N * cos(38°) = 100 N * 0.788 = 78.8 N
The vertical component, F_y, can be found by multiplying the force (100 N) by the sine of the angle (38°):
F_y = 100 N * sin(38°) = 100 N * 0.615 = 61.5 N
Step 2: Calculate the net force in the horizontal direction.
Since there is no friction, the only force acting in the horizontal direction is the applied force. Therefore, the net force in the horizontal direction is equal to the applied force:
F_net_x = F_x = 78.8 N
Step 3: Calculate the net force in the vertical direction.
In the vertical direction, the force of gravity (mg) acts downward and the force the ice exerts on the crate (F_up) acts upward. Therefore, the net force in the vertical direction is given by:
F_net_y = F_up - mg
Step 4: Calculate the acceleration of the crate.
Using Newton's second law of motion, we can find the acceleration (a) by dividing the net force in the horizontal direction (F_net_x) by the mass of the crate (m):
a = F_net_x / m = 78.8 N / 50 kg = 1.576 m/s^2 (rounded to three decimal places)
Step 5: Calculate the upward force exerted by the ice on the crate.
Using the net force in the vertical direction (F_net_y), we can solve for the upward force (F_up):
F_net_y = F_up - mg
F_up = F_net_y + mg
F_up = (F_y - mg) + mg
F_up = F_y = 61.5 N
Therefore, the acceleration of the crate is 1.576 m/s^2 and the upward force the ice exerts on the crate is 61.5 N.
To calculate the acceleration of the crate, we need to resolve the applied force into its horizontal and vertical components.
The horizontal component of the force can be found using the equation:
F_horizontal = F * cos(theta)
Where:
F = 100 N (applied force)
theta = 38° (angle with the horizontal)
Substituting in the values, we get:
F_horizontal = 100 N * cos(38°)
The vertical component of the force can be found using the equation:
F_vertical = F * sin(theta)
Substituting the values, we get:
F_vertical = 100 N * sin(38°)
Now, since there is no friction, the only horizontal force acting on the crate is the applied force, which is responsible for the acceleration. The equation relating force, mass, and acceleration is:
F = m * a
Solving for acceleration, we get:
a = F_horizontal / m
Substituting in the values of F_horizontal and m (mass), which is 50 kg:
a = (100 N * cos(38°)) / 50 kg
Now, let's calculate the acceleration:
a = (100 N * cos(38°)) / 50 kg
Next, to calculate the upward force the ice exerts on the crate, we need to consider the vertical forces acting on the crate. In this case, there are two forces: the vertical component of the applied force, and the force of gravity pulling the crate downward.
The force of gravity can be calculated using the equation:
F_gravity = m * g
Where:
m = 50 kg (mass of the crate)
g = 9.8 m/s^2 (acceleration due to gravity)
Substituting in the values, we get:
F_gravity = 50 kg * 9.8 m/s^2
Now, the upward force the ice exerts on the crate is equal in magnitude but opposite in direction to the vertical component of the applied force plus the force of gravity.
So the equation relating the upward force, vertical component of the applied force, and the force of gravity is:
Upward force = F_vertical + F_gravity
Substituting in the values of F_vertical and F_gravity, we get:
Upward force = (100 N * sin(38°)) + (50 kg * 9.8 m/s^2)
Now, let's calculate the upward force:
Upward force = (100 N * sin(38°)) + (50 kg * 9.8 m/s^2)