A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow
s: P is the upward force the woman exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the woman is trying unsuccessfully to lift the crate?
2 answers:
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Details are missing in the question. Complete text of the problem:
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"
Solution
(a) The pitcher accelerates the baseball from rest to a final velocity of , so
, in a time interval of
. The acceleration of the ball in the horizontal direction (x-axis) is therefore
And the distance covered by the ball during this time interval, before it is released, is:
(b) For this part we need to consider also the weight of the ball, which is
From this, we find its mass:
Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have
We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:
(upward)
So, the magnitude of the force is
To find the direction, we should find the angle of F with respect to the horizontal. This is given by
From which we find
Answer:
v_wind = 101.46 km / h , θ = 61.8
Explanation:
This is a velocity composition exercise.
Let's do the problem in parts. Let's start by knowing the speed of the plane without air.
v = d / t
v = 750 / 3.14
v = 238.85 km / h
This is the speed of the plane relative to the Earth and it does not change.
In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and
Let's use trigonometry to find the components of the plane's speed
cos θ = v_N / v
sin θ = v_W / v
v_N = v cos θ
v_W = v sin θ
let's calculate
V _N = 238.85 cos 22 = 221.46 km / h
v_W = -238.85 sin 22 = -89.47
the negative sign is because the plane is going west and the positive sign is the east direction.
As it indicates that the destination of the avine is towards the north, the x component of the wind must be
vₓ - v_W = 0
vₓ = v-w
vₓ = 89.47 km / h
in the direction to the East.
Now let's analyze the component of the wind in the Nort-South direction,
Indicate the travel time, let's calculate the speed that the component must have the speed of the plane
v_total = d / t
v_total = 750 / 4.32
v_total = 173.61 km / h
This is the final speed of the plane, which can be written
v_total = v_n - vy
vy = v_n - v_total
vy = 221.46 - 173.61
vy = 47.85 km
this component is directed towards the south
Let's use the Pythagorean Theorem, to find the magnitude
v_wind² = vₓ² + vy²
v_wind = √ (89.47² + 47.85²)
v_wind = 101.46 km / h
the address will then be found using trigonometry
θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹1 (47.85 / 89.47)
θ = 28.14
Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value
90- θtea = 90- 28.2
θ = 61.8
East of South