A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0º below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity?
(a) Alright, let's talk about this nurse and her cart. Picture a cart with groceries on it, being pushed by a nurse. We need to draw a free-body diagram to see what forces are acting on this cart.
First, we have the weight of the cart, which we'll represent with an arrow pointing straight down. Since the cart has a mass of 28.0 kg, let's give that arrow a magnitude of 280 N (mass multiplied by acceleration due to gravity, 9.8 m/s^2).
Next, we have the force of friction. Friction always opposes motion, so we'll draw an arrow pointing backward to represent this force. We know the force of friction is 60.0 N, so let's label the arrow with that value.
Finally, we have the force that the nurse exerts to move the cart. The nurse is pushing at an angle 35.0º below the horizontal, so let's draw an arrow at that angle. Since we don't know the magnitude of this force yet, we'll leave it blank for now.
(b) Now let's determine the force the nurse must exert to move the cart at a constant velocity. If the cart is moving at a constant velocity, that means the net force acting on it is zero. In other words, the force the nurse exerts must balance out the force of friction.
Since the force of friction is 60.0 N and it's pointing backward, the nurse must push with a force of 60.0 N in the opposite direction to balance it out. So the force the nurse must exert to move the cart at a constant velocity is 60.0 N, and that's no joke!
(a) The free-body diagram for the system of interest is as follows:
1. Gravitational force (mg): It acts vertically downward and has a magnitude of 28.0 kg × 9.8 m/s² (acceleration due to gravity) = 274.4 N.
2. Normal force (N): It acts perpendicular to the surface and has the same magnitude as the gravitational force, but in the opposite direction, which is vertically upwards. So, Normal force (N) = -274.4 N.
3. Force of friction (f): It acts horizontally in the opposite direction of motion (cart moving to the right). Its magnitude is given as 60.0 N.
4. Force exerted by the nurse (F): This force is applied at an angle of 35.0º below the horizontal. It can be resolved into two components: one along the horizontal axis (Fcosθ) and one along the vertical axis (Fsinθ).
┌───┐ N
│ │
│ │
f ← │ │ ← Fcosθ
│ │
└───┘
Fsinθ →
(b) To move the cart at a constant velocity, the force exerted by the nurse (F) must balance out the force of friction (f). This means Fcosθ = f.
Fcosθ = 60.0 N
F = 60.0 N / cos35.0º
Calculating this:
F ≈ 73.31 N
Therefore, the nurse must exert a force of approximately 73.31 N to move the cart at a constant velocity.
To determine the force that the nurse must exert to move the cart at a constant velocity, let's go step by step:
(a) Drawing a free-body diagram:
A free-body diagram helps us visualize and analyze the forces acting on an object. In this case, we want to draw the free-body diagram for the loaded cart. The forces acting on the cart are:
1. Weight (mg): It is acting vertically downward with a magnitude of (mass of the cart * acceleration due to gravity). Since the cart is on a horizontal surface, we can draw this force as a downward arrow.
2. Normal force (N): It is exerted by the ground on the cart and is perpendicular to the surface. Since the cart is on a horizontal surface, the normal force will be directed upward and equal to the weight of the cart.
3. Force of friction (Ffr): The frictional force opposes the motion of the cart and acts parallel to the surface. In this case, it has a magnitude of 60.0 N and will be directed opposite to the direction of the motion.
4. Force exerted by the nurse (F): This is the force that the nurse exerts on the handle of the cart. It acts at an angle of 35.0º below the horizontal.
(b) Calculating the force the nurse must exert to move at a constant velocity:
To find this force, we need to consider that the cart is moving at a constant velocity. This means that the net force acting on the cart must be zero since there is no acceleration.
The forces in the horizontal direction are the force of friction (Ffr) and the horizontal component of the force exerted by the nurse (Fcosθ).
Since the cart is moving at constant velocity, the frictional force must be equal in magnitude but opposite in direction to the horizontal component of the force exerted by the nurse:
Ffr = Fcosθ
Substituting the given values:
60.0 N = F * cos(35.0º)
To solve for F, we can rearrange the equation as follows:
F = 60.0 N / cos(35.0º)
Calculating this using a calculator, we get:
F ≈ 74.86 N
Therefore, the nurse must exert a force of approximately 74.86 N to move the cart at a constant velocity.
Fnet-Ff=ma
a=0
Fnet=Ff
Ff=Fnetcos(35)
Fnet=Ff/cos(35)
Fnet= 73N