a roller whose diameter is 20 in weight 72 lbs. what horizontal force is necessary to pull the roller over a brick 2 in high when (a) the force is applied to the center, (b) at the top

Interesting question.

the vertical height from the edge of the brick to the center is 18 inches. The horizontal distance from the edge of the brick to the center of gravity requires some calculation.

draw the figure. from the brick edge to the center is 20 inches. sketch a horizontal line frm the edge of the brick to the vertical downward radius. The You now have a triangle, 20 hypotenuse, 18 on the downward leg, and the horizontal leg will be d=sqrt(20^2-18^2)=sqrt(400-324)=sqrt 76 check that.

Now moments around the edge sum to zero.

F*18-weight*sqrt76=0
forceat center=72lbs*sqrt76/18

at the top, the process is similar.
Force*38-72sqrt76=0 solve for Force

(a) Well, pulling a roller over a brick can be quite a challenge, especially for the roller. If you apply the force to the center, it's like giving the roller a big bear hug to get it over the obstacle. So, let's calculate the horizontal force required.

To start, we need to convert the 20-inch diameter to a radius, which is half the diameter. Therefore, the radius is 10 inches. Now, let's add the 2 inches brick height, giving us a total of 12 inches.

To overcome this obstacle when applying the force at the center, the force needed can be calculated using the equation force = weight x (tan(angle of inclination) + coefficient of friction). Assuming a coefficient of friction of 0.3, we can proceed with the calculations.

The angle of inclination is given by the formula angle = arctan(height / radius).
angle = arctan(12 / 10) = arctan(1.2) ≈ 50.19 degrees.

Now, we plug this into the formula for the force:
force = 72 lbs x (tan(50.19 degrees) + 0.3).

After doing some math, we find that the horizontal force required to pull the roller over the brick when applying the force at the center is approximately X lbs. Well, let's hope the roller doesn't roll its eyes at our mathematical calculations!

(b) Now, this time the force is applied at the top. Yikes! Sounds like gravity is working against us, doesn't it? Well, let's crunch some numbers to see just how much force is needed.

When applying the force at the top, in addition to the weight, we need to counterbalance the gravitational force working on the roller. Since the force is applied at the top, the torque provided by the force must counterbalance the torque caused by the weight.

To calculate the horizontal force required in this case, it involves a bit more math. We'll need to use the equation force = (weight x radius) / (2 x height).

Plugging in our values, we get:
force = (72 lbs x 10 in) / (2 x 2 in).

After doing some calculations, we find that the horizontal force required to pull the roller over the brick when applying the force at the top is approximately Z lbs. So, my friend, it looks like this roller needs a little extra push when the force is applied at the top. Keep those muscles flexing!

To calculate the horizontal force required to pull the roller over a brick, we need to consider the weight of the roller and the angle at which the force is applied. Let's calculate the forces required in both scenarios:

(a) Force applied to the center:
In this case, the force will act directly opposite to the weight of the roller. The force required to overcome the weight can be calculated using the formula:

Force = Weight * tan(angle)

Given:
Diameter of the roller = 20 inches
Weight of the roller = 72 lbs
Brick height = 2 inches

First, let's convert the diameter to radius:
Radius = Diameter / 2 = 20 / 2 = 10 inches

Now, let's calculate the angle at which the force is applied:
Angle = arctan(Brick height / Radius) = arctan(2 / 10) = arctan(0.2)
Angle ≈ 11.31 degrees

Now, let's calculate the force required to pull the roller:

Force = 72 lbs * tan(11.31 degrees)

Using a calculator:
Force ≈ 14.23 lbs

Therefore, the horizontal force required to pull the roller over the brick when the force is applied to the center is approximately 14.23 lbs.

(b) Force applied at the top:
In this case, we need to calculate the normal force on the roller when it is at the top of the brick. The horizontal force required to overcome the weight and the normal force can be calculated as:

Force = (2 * Weight) * tan(angle)

Given:
Weight of the roller = 72 lbs
Brick height = 2 inches

Now, let's calculate the normal force at the top of the brick:

Normal force = Weight + (Weight * tan(angle))

Using a calculator:
Normal force ≈ 72 lbs + (72 lbs * tan(11.31 degrees)) ≈ 72 lbs + (72 lbs * 0.2) ≈ 72 lbs + 14.4 lbs ≈ 86.4 lbs

Now, let's calculate the force required to pull the roller:

Force = (2 * Weight) * tan(angle)

Using a calculator:
Force ≈ (2 * 72 lbs) * tan(11.31 degrees) ≈ 144 lbs * 0.2 ≈ 28.8 lbs

Therefore, the horizontal force required to pull the roller over the brick when the force is applied at the top is approximately 28.8 lbs.

To find the horizontal force required to pull the roller over a brick, we need to consider the weight of the roller, the height of the brick, and the distribution of force.

(a) When the force is applied to the center:
To calculate the required force, we need to consider the weight distribution and the principle of moments.

1. Determine the weight distribution:
Since the roller is cylindrical, the center of mass is in the middle of the roller's height. Therefore, the weight is evenly distributed on each half of the roller. Hence, each half weighs 36 lbs.

2. Calculate the moment of the weight:
The moment (torque) of an object is the product of its weight and the distance from the point of rotation (in this case, the center of the roller). Since the force is applied at the center, there is no moment to overcome.

Therefore, the required horizontal force to pull the roller over the brick when the force is applied to the center is zero. The weight of the roller alone is sufficient to provide the necessary force.

(b) When the force is applied at the top:
When the force is applied at the top, the roller will rotate around the point of contact with the brick. To determine the required force, we need to consider the moment generated by the roller's weight.

1. Calculate the moment of the weight:
The moment generated by the weight of the roller can be calculated using the formula:

Moment = weight * distance from the point of rotation

Since the force is applied at the top, the distance from the point of rotation is half the diameter of the roller, which is 20 inches / 2 = 10 inches.

Moment = 72 lbs * 10 inches = 720 lb-inches

2. Determine the force required:
To calculate the force required to overcome the moment, we use the equation:

Force = Moment / Distance from the point of force application

In this case, the distance from the point of force application is the height of the brick, which is 2 inches.

Force = 720 lb-inches / 2 inches = 360 lbs

Therefore, the horizontal force required to pull the roller over the brick when the force is applied at the top is 360 lbs.

the vertical height from the edge of the brick to the center is 8 inches. The horizontal distance from the edge of the brick to the center of gravity requires some calculation.

By illustration, from the brick corner to the center is 10 inches(since the given is diameter, divide it by 2 to get the radius which is equal to the distance from the corner of the brick to center of the roller). sketch a horizontal line from the corner of the brick to the vertical downward radius. Then now you have a triangle, 10 hypotenuse, 8 on the downward leg, and the horizontal leg will be d=sqrt(10^2-8^2)=sqrt(100-16)=6

Now moments around the edge sum to zero.

F*(10+8)-weight(6)=0
forceat center=72lbs

Force*(18)-72(6)=0 solve for Force

Answer is 24 lbs.