constant velocity and force?

Question

theres a garden roller which has a mass of 80kg
its pulled with a force of 250N at an angle of 30 degrees to the horizontal and it moves at constant velocity....

- how is the roller able to travel at constant velocity when being pulled by a force of 250N?
- if the roller was to be *pushed* with a force of 250N at an angle of 30 to the horizontal, what would be the changes that occur to the force exerted on the ground by the roller?

plzzzz help i am sooo stuck!!!
thanxxxx

Answer 1

Pulling:

The component of the 250N force in the vertical direction is 250 sin 30 (opposite to angle 30) and acts upward.

The weight of the roller, 80x9.8 N.

The net force acting on the roller is only
(80 x 9.8 - 250 sin 30) = 659 N, which acts down ward.

The weight is reduced.

The roller neither goes up or down.

Therefore, there must be a force acting upward on the roller which keeps it in equilibrium.

This force is the normal reaction acting upward and equals 659 N.


The horizontal force acting on the roller is 250 cos 30
(Adjacent to 30 degree) = 216.5 N.

Since the roller moves with constant speed, the net force must be equal to zero.

(Note that when the net force is zero the body will be either at rest or moves with constant speed. In fact, we need not speak about rest and it is enough if we say that the body moves with constant speed. It is because; rest means a constant speed of zero.)

There must be a force opposing the force 216.5 N.

This force must be the frictional force 216.5 in the opposite direction.

We can find the coefficient of friction since we know the normal reaction is 659 N.

The coefficient is 216.5/ 659 = 0.33.

Pushing.

The component of the 250N force in the vertical direction is 250 sin 30 (opposite to angle 30) and acts downward.

This increases the total downward force to

250 sin 30 + 80 x 9.8 = 909N.

Since the roller neither moves up or down there must be a force in the upward direction and it is the normal reaction = 909N.

The horizontal force = 250 cos 30 = 216.5 N.




As it is found that when pulling with 250 N forces the roller moved with constant speed, the coefficient of friction calculated using that will not change.

Now the frictional force will be µ x normal reaction

= 0.33 x 909 = 300 N (rounded to)

But this is the maximum frictional force.

Since this is more than the applied force 216.5, the actual frictional force will be only 216.5 N and since this is less than the limiting force the roller will not move till we give a horizontal force of 300 N.

THE ROLLER IS AT REST.

We conclude more force is needed when pushing than pulling with the same force in magnitude and direction.

Answer 2

Your both queries are genuine. Yes, you are right if roller were to be pulled only by 250 N force it would not travel with constant velocity. The very fact it has been told that it is moving with constant velocity, means there is some other force acting to oppose this force. What that can be, you have to guess. But if it has not been given does not mean that you have been told that it does not exist. In this situation you have to conclude that it must be the ground which must be applying some force. Now in which direction ground can apply force. Only two dircetions normal to it and tangential to it. Before you started pulling the roller, it was standing there merrily even though earth was pulling it towards the centre. This could happen because ground is hard it offers resistance to change of its shape just as a spring does. But what happens in spring is visible, here it is not visible. As a rsult of this ground applies equal and opposite force on the roller in opposite direction. This is called Normal reaction.. Rememebr that this is not fixed but always adjust to the situation. Now if you push the roller horizontaly as gently as possible. You are applying a force but the roller does not move, that means ground is also capable of applying tangential force, popularly known as frictional force. This force is also self-adjusting and it is always equal to the applied force when the body is permanently at rest or moving in a straight line with constant velocity.

Here comes your next query. The force of 250 N can be thoght as made of two forces 250xsin30 = 125 N along horizontal direction and 250xcos30 = 250x0.86 = 215 N along vertically upward direction. This reduces the force acting on the roller towards the centre of earth by 215 N and hence also reduces the self-adjusting normal reaction by the ground. So now it is obvious that the tangental frictional force which the ground is applying on the roller in the opposite direction of its motion is 125 N. The maximum frictional force which a surface can apply on a moving or tending to be moving object, called as limiting friction is proportional to the normal reaction. By pulling the roller instead of pushing it we reduce the normal reaction and hence the force of limiting friction. Your further queries are welcome

Answer 3

Q1: Friction equals the horizontal component of the pulling force. So, the NET horizontal force is zero → a = 0.0

Q2: When you push, you increase the downward force on the roller, also known as Fn. Since The frictional force equals µ*Fn, Ff increases. So, both the x and y component of the roller's force on the ground increase compared to pulling it.

That's why it's usually easier to pull something heavy than to push it...........

Answer 4

It means that balance of forces on roller =0; thus let us verify that 80*g*sin(30°)=250N; 80*9.8*(1/2)= 392.4>250; no! the roller would slip down no matter pulled or pushed.
The force exerted = 80*9.8*cos(30°) = 680N;

Answer 5

when it travel at a constant velocity, that means all the forces are balance.

Friction is equal to Fapply.

It doesn't really matter if you push or pull.
what matter is the direction you going.
Think about the "pulling". You pull with an angle but you go horizonally.

Answer 6

The friction is not equal to Fa in this case because there is a angle.

Friction should be equal to the Fa(x) which is 216.5N