Answer:
the normal force the ground exerts on the crate is 1470 N, and the applied force of the rope is 700 N.
Step-by-step explanation:
To solve this problem, we can use the equations of motion and Newton's laws of motion.
First, let's consider the forces acting on the crate:
The force of gravity, which is equal to the mass of the crate multiplied by the acceleration due to gravity (Fg = mg). This force is downward.
The normal force, which is the force exerted by the ground on the crate perpendicular to the ground (FN). This force is also downward.
The force of friction, which is the force that opposes the motion of the crate and is equal to the coefficient of friction times the normal force (Ff = μFN). This force is opposite to the direction of motion.
The applied force of the rope, which is the force that is pulling on the crate at an angle of 36° above horizontal (Fa). This force is upward and to the right.
Now, let's apply Newton's second law of motion to the crate. The sum of the forces acting on the crate in the horizontal direction (Fa - Ff) must be equal to the mass of the crate multiplied by the acceleration in the horizontal direction (ma). The sum of the forces acting on the crate in the vertical direction (Fg - FN) must be equal to the mass of the crate multiplied by the acceleration in the vertical direction (ma).
We can use these equations to solve for the normal force and the applied force of the rope.
First, let's solve for the normal force:
Fg - FN = ma
FN = Fg - ma
FN = (150 kg)(9.8 m/s^2) - 0
FN = 1470 N
Now, let's solve for the applied force of the rope:
Fa - Ff = ma
Fa = Ff + ma
Fa = (700 N) + 0
Fa = 700 N
Therefore, the normal force the ground exerts on the crate is 1470 N, and the applied force of the rope is 700 N.