a roller whose diamater is 1.0m.weight 360N.what horizontal force is necessary to pull the roller over a brick 0.1m,heigh when the force is applied at the centre?
To calculate the horizontal force necessary to pull the roller over the brick, we need to consider the gravitational force acting on the roller and the work done to lift it over the brick.
First, let's calculate the gravitational force acting on the roller. The weight of the roller is given as 360 N, which represents the force pulling it down vertically. Since the roller is on a flat surface, this force acts vertically downward. We will take the positive x-direction as horizontal, so there is no component of the weight in that direction.
Next, let's calculate the work done to lift the roller over the brick. Work is defined as force multiplied by displacement, and in this case, the force is applied at the center of the roller, so we can assume the force acts on a point directly opposite the brick. The displacement is given as the height of the brick, which is 0.1 m.
The work done is given by the formula:
Work = Force * Displacement * cos(theta)
Since the force and the displacement are both in the x-direction, the angle between them (theta) is 0 degrees, and cos(0) = 1. Therefore, we can ignore the cosine term in this case.
The work done to lift the roller over the brick is:
Work = Force * Displacement
Since the work done is equal to the change in gravitational potential energy, we have:
Work = m * g * h
where m is the mass of the roller, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the brick.
Let's rearrange this equation to solve for the force:
Force * Displacement = m * g * h
Force = (m * g * h) / Displacement
Now we need two additional pieces of information: the mass of the roller and the distance over which the force is applied.
If we assume the roller is a uniform disk, we can use the formula for the moment of inertia of a disk:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass, and r is the radius of the roller. Since the diameter of the roller is given as 1.0 m, the radius is 0.5 m.
Now, let's calculate the mass of the roller using the weight given:
Weight = m * g
m = Weight / g
Substituting the values:
m = 360 N / 9.8 m/s^2 ≈ 36.73 kg
Next, let's calculate the moment of inertia I:
I = (1/2) * m * r^2
I = (1/2) * 36.73 kg * (0.5 m)^2
I ≈ 4.59 kg·m^2
Now we have all the necessary information to calculate the horizontal force:
Force = (m * g * h) / Displacement
Force = (36.73 kg * 9.8 m/s^2 * 0.1 m) / 1.0 m
Force ≈ 35.94 N
Therefore, a horizontal force of approximately 35.94 Newtons is necessary to pull the roller over the brick when the force is applied at the center.
To determine the horizontal force required to pull the roller over a brick of certain height, we need to consider the weight of the roller and the height difference.
Given data:
Diameter of the roller (d) = 1.0 m
Weight of the roller (W) = 360 N
Height of the brick (h) = 0.1 m
When the force is applied at the center of the roller, it results in a torque that contributes to rotational motion. However, since the roller is not rotating in this scenario, the torque produced by the weight of the roller must be balanced by an opposing torque.
The torque produced by the weight of the roller can be calculated as follows:
Torque = weight x radius
To calculate the radius, we divide the diameter by 2:
Radius (r) = d/2 = 1.0 m/2 = 0.5 m
Torque = 360 N x 0.5 m = 180 N.m
To balance this torque, there must be an opposing torque exerted by the horizontal force. The opposing torque can be calculated using the equation:
Torque = Force x Distance
In this case, the distance is the height of the brick, h = 0.1 m. Therefore:
180 N.m = Force x 0.1 m
Rearranging the equation to find the horizontal force:
Force = Torque / Distance = 180 N.m / 0.1 m = 1800 N
So, a horizontal force of 1800 N is required to pull the roller over a brick that is 0.1 m high when the force is applied at the center.