Anyone know anything about physics?

Question

a 100kg crate is resting on level ground, with a cluster of helium balloons tied to the top. The helium balloons provide an upward force of 100 N.
a) Draw a freebody diagram of the forces on the crate.
b) Calculate the force of gravity on the crate.
c) If you must apply a 1000 N force to get the crate to move, what is the coefficient of static friction newt(s)?
d) If you apply a horizontal 1100 N force to the box, what will be its rate of acceleration?(consider friction)

Answer 1

b)
The force of gravity on the crate = 100 x10 = 1000 N.
c).
The net downward force = 1000 - 100 = 900 N
The normal reaction by the ground upward = 900 N
μ = Normal reaction / frictional force.
Since the crate is at the verge of motion the applied force equals the maximum frictional force. The frictional force = 1000 N.
μ = 900 / 1000 = 0.9

d ]
The net force acting horizontally on the crate = 1100 - 1000 =100 N.
Acceleration = force / mass = 100 / 100 = 1m/s^2.

a) for free body diagram .
Represent the box by a point O.
Draw to scale a 100 N force upward.
100 x10 =1000N downward.
Draw normal reaction force 900 N upward.

Answer 2

a) Draw a freebody diagram of the forces on the crate.
B
^
|
U
|
V
W

Where U is the crate, B is buoyancy the up force from the balloons = 100 kg m/sec^2, and W = mg = 100 * 9.81 ~ 98 kg m/sec^2 which is the weight of the crate (force of gravity down) of mass m = 100 kg.

b) Calculate the force of gravity on the crate.

Did that above, W ~ 98.1 kg m/sec^2 (aka Newton)

c) If you must apply a 1000 N force to get the crate to move, what is the coefficient of static friction newt(s)?

This does not apply because f = ma = B - W > 0; so that a > 0 and the balloon is rising off the ground. As it is not touching the ground there is no contact and coefficient of friction is meaningless. [Which is why one answer got a coefficient > 1.0, which is physically impossible.]

d) If you apply a horizontal 1100 N force to the box, what will be its rate of acceleration?(consider friction)

As there is no friction while the crate is in the air, f = ma = P; where P = 1,100 kg m/sec^2 so that a = P/m ~ 1.1 m/sec^2.

I'm wondering where you got this question? If there was an answer for c), then you must have left something out or miscopied something because when the crate is lifting off the ground friction force F = kN; where N = B - W > 0 means the normal force is acting upward, not downward onto the ground. So there is no friction force. Ditto for d), no friction force when the crate is in the air.

Answer 3

I can't draw a diagram in here.
Well, first let's calculate the force of the crate toward the ground. This is 100kg*[gravitational acceleration] = X newtons, 100kg*9.81=981N. (force of gravity on the crate)
So, there is an upward force of 100N, and a downward force of 981N, giving a NET downward force of 981-100=881N. The coefficient of static friction times the downward force will be equal to the horizontal force necessary to move the crate, in other words where muS is the coefficient of static friction, 881N*(muS)=1000N -> muS = 1000/881 = 1.14. (coefficient of static friction).
Ok, so muK is the coefficient of static friction, the forces on the crate are (downward) 881N, (horizontal) 1100N, (negative horizontal) muK*881N. The acceleration can be given by taking the force (newtons, kg*m/s^2) and dividing it by the mass (kg), where the force is going to be 1100N-(muK*881N) and the weight is 100kg, so (1100N-muK*881N)/100kg = [1.1 - (muK*.881)] m/s^2
As a side note, the acceleration is a linear function of the coefficient of kinetic friction.

Answer 4

Arrow up 100N. Arrow down = mg = 1000N = Fgravity

Ffriction = mu Fnormal so 1000 = mu (900) so mu = 1.1

weird because usually mu is less than 1 !!!!

Fnet = 1100 - 1000 = ma = 100 = 100a so a = 1m/s^2

This is not hard !!!

Answer 5

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