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if a ramp is at a 45 degree angle by waht percent will a 1kg differ from freefall and going on a ramp downward. note gravity is 9.8m/s^2, what about a 30' angle ramp? where/what is a formula

2006-10-17 04:55:05 · 9 answers · asked by qwe 3 in Science & Mathematics Physics

9 answers

SIN(angle)

SIN(45) = .7, so any mass accelerates at .7*9.8=6.9m/s^2
SIN(30)=.5 so any mass accelerates at .5*9.8=4.9m/s^2.
SIN(0)=0 so any mass just sits there on a flat surface since 0*9.8=0 acceleration.

2006-10-17 05:07:44 · answer #1 · answered by campbelp2002 7 · 0 1

The change in the direction of weight (W = mg) with respect to the surface of the ramp is W cos(deg) = mg cos(deg); where deg is measured between the horizontal and the slope of the ramp. W cos(deg) = N = normal weight because its vector is perpendicular (normal) to the ramp's surface.

We can easily verify this by assuming the ramp is flat on the ground; therefore, deg = 0 and cos(0) = 1.000. So the normal weight = N = W, which is intuitive in that the normal weight and basic weight are the same weights. Conversely, if deg = 90, then cos(90) = 0 and there is no weight on the ramp from the M = 1 kg mass.

I address normal weight (N) because it is important in Ff = kN; where Ff is the force of friction between your mass and the ramp, k is the coefficient of friction, and N is the normal weight acting on the ramp. The friction force acts to slow down a sliding body on a ramp and the sliding weight (S), which is the weight vector pointing along the ramp, acts to speed it up.

Thus, f = Ma = S - Ff; where f is the net force acting on the mass (M) along the ramp, S = W sin(deg), and Ff = kN = k Wcos(deg) Note the sliding weight that pulls the mass M down the ramp varies according to the sin of the angle between the ramp and the horizontal.

We can test this relationship by assuming "free fall," which will occur when deg = 90 so that the ramp no longer supports the body. In which case, f = Ma = W(sin(deg) -kcos(deg)) = W = Mg or, in free fall, the 1 kg mass is falling at g = 9.81 m/sec^2, the acceleration due to gravity at the Earth's surface. This is exactly the result one would expect of a free falling body near the Earth's surface.

For sliding down the ramp, we assume angles less than 90 deg with the horizontal. For example, f = Ma = W[sin(45) - k cos(45)] = Mg[sin(45) - k cos(45)] when solved for (a) will give you the acceleration of the mass M down a 45 deg slope. You can work this problem for the degrees you gave if you have k, the coefficient of sliding friction, which you did not give.

Without k, you cannot figure the friction force that counteracts the weight vector down the ramp's surface. But even so, by examining the equation above you can see how the acceleration down the ramp will be affected by the angle of the ramp.

2006-10-17 05:49:17 · answer #2 · answered by oldprof 7 · 0 0

To clarify--ramps (or any other physical structure smaller than a large asteroid) has no effect on gravity. However, as was pointed out about, there will be an effect on the resultant acceleration. Since the "rider" on the ramp can't move in the direction of gravity ("down"), to determine the actual force acting in the direction of movement, you use the equation Force = g * sin (A), where g = 9.8 m/s^2 and A = the angle of the ramp. You can see this if you draw a force-body diagram, and remember that gravity (as all forces) is a vector--that is, it has both direction and magnitude. The same general principle is used whenever the force on an object is not in the same direction as the motion--a simlar problem would be a child pulling a wagon, or wind blowing a sailboat.

2006-10-17 05:08:57 · answer #3 · answered by Qwyrx 6 · 0 0

"if a ramp is at a 45 degree angle by waht percent will a 1kg differ from freefall " You need to state the question clearly, what are you looking for? Acceleration?

2006-10-17 05:06:16 · answer #4 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0 0

w=f(cos(theta))(d) you cant have a ramp of ninety tiers in any different case there is not any gravity pushing the object onto the ramp and the ramp isn't doing something In a perfect type the perspective nevertheless effects the artwork accomplished as a results of fact gravity impression's the object in yet differently diverse quantities of the products finished gravity are pushing the gravity to the ramp

2016-12-08 16:13:18 · answer #5 · answered by money 4 · 0 0

A]Free fall Weight=mg
=1kg*9.8m/sec62=9.8Newtons
B]Comin down the ramp the mass is
constrained to move along the ramp
Hence only a component of weight
will come in to play: thus
force acting along ramp
=mass*acceleration*sin45Deg
=mg*1/sqrt2
=.707 mg
There is nearly 70% reduction
C]If the ramp angle is 30Deg
force=mg*sin30
=1/2 mg
There is 50% reduction

2006-10-17 05:37:07 · answer #6 · answered by openpsychy 6 · 0 0

When on a ramp, force pointing in the vertical direction are broken down into their components. The force due to gravity is mg. When the mass is 1kg, the force of gravity is just "g." When g is broken down into its components, you have a force in the horizontal direction and a force in the vertical direction:

gx = g*sin(θ)

gy = g*cos(θ)

θ is whatever angle the ramp is elevated at.

As you can see, the component forces are smaller than the net force, due to the restricted domains of the sine and cosine functions.

2006-10-17 05:05:36 · answer #7 · answered by عبد الله (ドラゴン) 5 · 0 0

Fp (force parallel to the ramp) =WsinA
Fn (force normal to the ramp) = WcosA
where w = weight and A is the angle of the ramp.
g = W/m, a = F/m, so
an = gcosA
ap = gsinA

ap/g = sinA

sin 45 = .707 = 70.7%
sin30' = 0.087 = 8.7%

2006-10-17 06:19:26 · answer #8 · answered by Helmut 7 · 0 0

g sin 45

2006-10-17 05:02:07 · answer #9 · answered by theliberaloneforyou 1 · 0 0

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