A lawn roller of mass 200kg is being push at constant speed by handle,which is incline at 40°to the horizontal.If the coefficient of friction 0.12,calculate the force be applied to handle by pusher?

To calculate the force that needs to be applied to the handle by the pusher, we need to consider the forces acting on the lawn roller.

The forces acting on the lawn roller are:
1. Weight of the roller acting downwards (mg)
2. Normal force acting perpendicular to the incline (N)
3. Force of friction opposing the motion (f)
4. Force applied by the pusher (F)

Since the roller is moving at a constant speed, the net force acting on it is zero. Therefore, the force applied by the pusher is equal in magnitude and opposite in direction to the sum of the weight component parallel to the incline, the force of friction, and the component of the normal force parallel to the incline.

The weight component parallel to the incline is given by:
mg*sin(40°)

The normal force is equal in magnitude and opposite in direction to the weight component perpendicular to the incline, which is given by:
mg*cos(40°)

The force of friction is given by:
f = μ*N = 0.12*N

Setting up the equation for the net force:
F = mg*sin(40°) + 0.12*mg*cos(40°)

Substitute the values:
F = 200*9.81*sin(40°) + 0.12*200*9.81*cos(40°)
F = 1316.8 + 101.6
F = 1418.4 N

Therefore, the force that needs to be applied to the handle by the pusher is 1418.4 N.

Free Body Diagram:


Given:

- Mass of the lawn roller, \(m = 200\text{ kg}\)
- Angle of the handle, \(\theta = 40^\circ\)
- Coefficient of friction, \(\mu = 0.12\)
- Acceleration of the lawn roller, \(a = 0 \text{ m/s}^2\) (constant speed)

Solution:

Vertical forces:

$$N - mg\cos\theta = 0$$

Horizontal forces:

$$F - mg\sin\theta - f = 0$$

where \(f\) is the friction force.

Friction force:

$$f = \mu N$$

Substituting (3) into (2):

$$F - mg\sin\theta - \mu(mg\cos\theta) = 0$$

Solving for \(F\):

$$F = mg(\sin\theta + \mu\cos\theta)$$

$$F = (200\text{ kg})(9.81 \text{ m/s}^2)(\sin40^\circ + 0.12\cos40^\circ)$$

$$F \approx \boxed{390.6 \text{ N}}$$

Therefore, the force that must be applied to the handle by the pusher is approximately 390.6 N.