Rachmaninov's lament

The minimum force required is

$\mathrm{mg\sin\theta}$ = $\mathrm{ 500 * 9.8 * \sin\ 30^ \circ}$

= 500 * 9.8 * $\frac{1}{2}$

= 2450 newtons

We know that, for ramped planes,

$a=g \sin \theta = 9.8 m s^{-2} \times \sin 30∘ = 9.8 ms^{-2} \times 0.5 = 4.9 ms^{-2}$

Now, we have $m=500 kg$ and $a=4.9 ms^{-2}$ ... Let's plug in into $F=ma$ ...

$F= 500 kg \times 4.9 ms^{-2} =2450 N$ ...

Hence, the answer is $\fbox{2450N}$ ...

Taking friction to be negligible, there are only 2 forces acting on the piano. Weight and contact force. We shall take the vertical direction to be perpendicular to the slope and the horizontal to be parallel to the slope. You can resolve weight into its components - in the vertical direction, Weight and Contact force cancel each other out. With the horizontal component, W sin(30), we find the amount of force pushing the piano down the ramp. This is the only force we need to care about as the vertical forces are negligible. To overcome the force of W sin(30), your minimum force must also be W sin(30). mg sin(30) = (4900)(1/2) = 2450 N

The object (concert grand piano) receives 2 distinct force.

1. Supporting force from the ground, $P$ .

2. Gravity force $W$ .

We need to find the force to push the object up.

Now, we solve for the gravity force, which is

$mg$

$=500kg\times9.8m/s$

$=4900N$

Now, we solve for the force that pull the object down (direction is parallel to the ground, Let it be $R$ ). Note that there is only gravity force could effect on the force pulling the object down. Therefore, we can solve it through the law of triangle of forces.

$W$ is the hypotenuse; $R$ is the adjacent side; included angle is $60^\circ$

$R=Wcos60^\circ$

$R=4900cos60^\circ$

$R=2450N$

To push the object up, the force must be larger than $2450N$ , so $2450N$ is the minimum force.

The formula used is mg $\sin \theta$

So mg* $\sin \theta$ = 500 * 9.8* $\sin 30$

Answer = 2450

Found the force needed in the horizontal of the ramp to make the piano to stay stop that is the sin30xWeight. The force needed is higher to 2450. So, 2451.

The force of gravity is always straight down, but because of the ramp the force will be distributed into two new x and y components, which we can call x' and y'. We can set x' as the component parallel to the line, and y' as the component perpendicular to the line.

We also know that there is a normal force that prevents the piano from passing through the ramp, and this force will equate the y' component of the weight force. All that is left from the weight force is the x' component, which we can compute.

Because we essentially rotated the x/y axis by 30 degrees counter-clockwise to obtain x' and y', to separate the gravity into these components we take the force of gravity and then turn them into components.

The y' component will be (500kg) (-9.8m/s^2) cos(30 deg), which will be canceled by the normal force.

The x' component will be (500kg) (-9.8m.s^2) sin(30 deg), which is the force we need to push to counteract the gravity force.

The angular version of $F=m \times a$ here would be $mg \sin \theta$ where use g will be made into 9.8. We fill in these variables with the given information to give $m\times g \sin \theta=500\times\ 9.8\times \sin 30= 500\times9.8\times\frac{1}{2}$

BC/a = cos 30 BC = a x cos 30 9.8 x 0.5 = 4.9 mass = 500 kg x 4.9 = 2450

f=mgsin(theta) =500 9.8 sin(30) =2450

We are given with the following data:

• mass M = 500 Kg
• angle of inclination \theta = 30 degrees

We need to find the minimum amount of force in Newtons (N) . According to Newton's First Law of Motion, an object tends to be at rest or at constant velocity (constant motion, no speeding up or slowing down), unless an external force is applied to it. Commonsensically, the object would slide down the ramp unless something else stops it. So, in order for us to keep it at rest or at uniform motion, we must counter-act that force that causes it to slide.

Let's assume that the ramp is a right triangle, with the angle of inclination of 30 degrees wrt the horizontal. Any object that has mass also has weight. And since weight is a vector quantity and a kind of force, it must have a direction. This direction is what it would take given the force. Weight always points toward the center of Earth, so in this case, it should be perpendicular to the ground.

Here goes the ramp! The mass also applies a force on the ramp, since the latter kind of "blocks" the object's original path if it was not there. Remember that the ramp is inclined, so this must have an effect on why the object slides down. Actually, the object slides down because of its own weight and the angle that ramp makes wrt the ground. In this case, the force F vector is actually divided into two coordinates, one that points to bottom part of the ramp, and the other one perpendicular to and is on the plane of contact between the object and the ramp. We only need the first one:

$\sin \theta \times M \times g =$ force $F$ needed to push the object up

$\sin 30 \times 500 \times -9.8 = \frac{1}{2} \times 500 \times -9.8$

$= 2450$ Newtons

Force required to push a body at a gradient is given by F =m x g x sin(theta). For 30 degree gradient, value of sin 30 is 0.5. Therefore, force required is 500 x 9.8 x 0.5 = 2450 N

F = mg sin @

F = (500)(9.81) sin (30)

F = 2452.5 N

force in Newtons need to push a m kg thing up a ramp angled @ is mgsin@.

The force required to push the concert grand piano up a ramp angled 30° with respect to the horizontal is $500\cdot9.8sin(30°)$ which is equal to 2450 Newtons.

We know Fs= µsR where µs=tan⁡θ=sin⁡θ/cos⁡θ and R=mg cos⁡θ ∴ Fs=sin⁡θ/cos⁡θ × mg cos⁡θ=mg〖 sin〗⁡θ Given M=500kg and θ=30° ∴ Fs=500kg×9.8ms^-2×〖 sin〗⁡〖30°〗 =2450N (Ans)

To find the minimum force needed to push the piano up the ramp you need to determine the component of gravity that is pulling the piano down the ramp. To do this you can draw a FBD with the x-axis aligned with the ramp. After drawing the x-component of gravity you will realize that it is mgsin(30), which gives you your answer of 2450 .

F=P ; como ele quer a força que atua na horizontal temos a Fx, ou seja, Px.

Utilizaremos o seno de 30º= 1/2 ou 0.5

Sen30º=Px/P

Px= P*sen30º

Px= 500*9.8 / 2

Px=2450

If we consider that there is no friction in the ramp, the minimum force is :

F = M.g.sen@ ---> F = 500 kg . 9,8m/s² . 0.5

F = 2450 N