A 40 kg crate is pulled across the ice with a rope. A force of 90 N is applied at an angle of 36 degreees with the horizontal. Neglecting friction, calculate: the acceleration of the crate; the upward force the ice exerts on the crate as it is pulled
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To answer your question, we need to break down the force applied to the crate into its horizontal and vertical components.
First, let's find the horizontal component of the force applied. We can do this by multiplying the magnitude of the force (90 N) by the cosine of the angle (36 degrees).
Horizontal component = 90 N * cos(36°)
Next, let's find the vertical component of the force applied. We can do this by multiplying the magnitude of the force (90 N) by the sine of the angle (36 degrees).
Vertical component = 90 N * sin(36°)
Since there is no friction mentioned, the net horizontal force acting on the crate will be responsible for its acceleration.
The formula for net force is given by Newton's second law of motion, which states that the net force (F_net) is equal to the mass (m) multiplied by the acceleration (a).
F_net = m * a
In this case, the net force equals the horizontal component of the force applied. So we can write:
90 N * cos(36°) = 40 kg * a
Now we can solve for the acceleration:
a = (90 N * cos(36°)) / 40 kg
Using a calculator, we find that the acceleration of the crate is approximately 1.23 m/s^2.
Next, let's address the upward force that the ice exerts on the crate. Since there is no vertical acceleration, it means the sum of the vertical forces is zero.
The upward force exerted by the ice is equal in magnitude but opposite in direction to the vertical component of the force applied.
Upward force by ice = - (90 N * sin(36°))
Using a calculator, we find that the upward force exerted by the ice is approximately -51.6 N. Note that the negative sign indicates that the force is acting in the opposite direction to the downward force of gravity.