For a winter fair, some students decide to build 30.0- kg wooden pull-carts on sled skids. If two 90.0- kg passengers get in, how much force will the puller have to exert to move the pull-cart? The coefficent of maximum static friction between the cart and the snow is 0.15.
3.1•10^2
Well, well, winter fair shenanigans! Let's calculate the amount of force required to make things slide.
First, let's find the maximum static friction force between the sled and the snow. We can do that by multiplying the coefficient of static friction (0.15) by the weight of the sled (30.0 kg).
Frictional force = coefficient of static friction × weight
Frictional force = 0.15 × 30.0 kg
Now, since the sled is carrying two 90.0 kg passengers, we need to account for their weight too. So, the total weight being pulled is:
Total weight = weight of sled + weight of passengers
Total weight = 30.0 kg + (2 × 90.0 kg)
Finally, the force the puller needs to exert is the sum of the frictional force and the total weight:
Force = frictional force + total weight
Go ahead and crunch those numbers, my friend, and let me know what you get.
To determine the force the puller must exert to move the pull-cart, we need to consider the forces acting on the cart.
1. Weight of the pull-cart (mg):
The weight of the pull-cart is the mass (m) multiplied by the acceleration due to gravity (g).
Weight of the pull-cart = m * g
Given:
Mass of the pull-cart (m) = 30.0 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Plugging in the values:
Weight of the pull-cart = 30.0 kg * 9.8 m/s^2
2. Weight of the passengers:
The weight of the passengers is the mass (m) multiplied by the acceleration due to gravity (g).
Weight of the passengers = m * g
Given:
Mass of each passenger (m) = 90.0 kg
Plugging in the values:
Weight of the passengers = 90.0 kg * 2 * 9.8 m/s^2 (since there are two passengers)
3. Maximum static friction force (fs):
The maximum static friction force is determined by the coefficient of maximum static friction (μs) multiplied by the normal force (N).
fs = μs * N
Given:
Coefficient of maximum static friction (μs) = 0.15
The normal force (N) can be calculated as the sum of the weight of the pull-cart and the weight of the passengers.
N = Weight of the pull-cart + Weight of the passengers
Plugging in the values:
N = (30.0 kg * 9.8 m/s^2) + (90.0 kg * 2 * 9.8 m/s^2)
Now we can calculate the maximum static friction force (fs) using the coefficient of maximum static friction (μs) and the normal force (N).
4. Force exerted by the puller (F):
The force exerted by the puller needs to overcome the maximum static friction force (fs) to set the cart in motion.
F = fs
Plugging in the values:
F = 0.15 * (30.0 kg * 9.8 m/s^2 + 90.0 kg * 2 * 9.8 m/s^2)
Now you can calculate the force (F) exerted by the puller to move the pull-cart.
To find the force that the puller will have to exert to move the pull-cart, we can use Newton's second law of motion, which states that the force (F) exerted on an object is equal to the mass (m) of the object multiplied by its acceleration (a). In this case, the acceleration will be the acceleration required to overcome the force of friction.
First, let's calculate the force of friction between the wooden pull-cart and the snow. The force of friction can be determined using the equation:
Frictional force (Ffriction) = coefficient of friction (μ) * normal force (N)
The normal force (N) acting on the pull-cart is equal to the weight of the pull-cart and the passengers, which can be calculated as:
Normal force (N) = (mass of pull-cart + mass of passengers) * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2.
Let's plug in the given values:
Mass of pull-cart (m1) = 30.0 kg
Mass of passengers (m2) = 90.0 kg
Coefficient of maximum static friction (μ) = 0.15
Acceleration due to gravity (g) = 9.8 m/s^2
First, calculate the normal force:
Normal force (N) = (30.0 kg + 90.0 kg) * 9.8 m/s^2
Normal force (N) = 120.0 kg * 9.8 m/s^2
Normal force (N) = 1,176 N
Next, calculate the force of friction:
Frictional force (Ffriction) = 0.15 * 1,176 N
Frictional force (Ffriction) = 176.4 N
Now we have the force of friction acting against the puller. To overcome this force, the puller must exert an equal and opposite force, so the force exerted by the puller (Fpuller) will also be 176.4 N.
Therefore, the puller will have to exert a force of 176.4 N to move the wooden pull-cart with two 90.0 kg passengers on it.