(a) Find the single force that is equivalent to the two forces shown. The forces
ID: 1659582 • Letter: #
Question
(a) Find the single force that is equivalent to the two forces shown. The forces are measured in units of newtons (symbolized N).
N î + N
(b) Find the force that a third person would have to exert on the mule to make the resultant force equal to zero.
N î + N
2.)A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.10 m/s at an angle of 15.0° below the horizontal. It strikes the ground 6.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
m
(b) Find the height from which the ball was thrown.
m
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
s
3.)A basketball player is standing on the floor 10.0 m from the basket as in the figure below. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, = 44.0°, with the horizontal from a height of h = 1.86 m.
(a) What is the acceleration of the basketball at the highest point in its trajectory?
(b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
m/s
Explanation / Answer
IN vector notation
force on right F1 = 90.1*cos50 i + 90.1*sin50 j
force on right F1 = 57.9 i + 69.0 i
force on left F2 = -79.7*cos60 i + 79.7*sin60 j
F2 = -39.8 i + 69.0 j
single force that is equivalent to the two forces F = F1 + F2
F = -18.1 N i + 138 N j <<<<<<------ANSWER
==================================
(b)
resultant R = F1 + F2 + F3
given resultant R = 0
F1 + F2 + F3 = 0
F + F3 = 0
F3 = -F
F3 = 18.1 i - 138 j
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2)
(a)
along horizontal
initial velocity vox = vo*costheta
aceeleration ax = 0
time interval t = 6 s
displacement = x
from equation of motion
x = vox*t + (1/2)*ax*t^2
x = (8.18cos15)*6 + 0
x = 47.4 m <<<<_____________ANSWER
(b)
along vertical
initial velocity voy = -vo*sintheta
acceleration ay = -g = -9.8 m/s^2
initial point y0 = h
finalpoint y = 0
time t = 6 s
from equation of motion
y-y0 = voy*t + (1/2)*ay*t^2
-h = -(8.1*sin15*6) - (1/2)*9.8*6^2
h = 189 m
(c)
final point y= 10 m
10 - 189 = -(8.1*sin15*t) - (1/2)*9.8*t^2
t = 5.83 s <<<<-------ANSWER
==========================
(3)
the only force acting is gravitational acceleration = -g = -9.8 m/s^2
magnitude = 9.8 m/s^2
direction = vertically down wards
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(b)
PROJECTILE
1)
along horizontal
________________
initial velocity v0x = v*cos(theta)
acceleration ax = 0
displacement x = 10 m
from equation of motion
x = v0x*T+ 0.5*ax*T^2
x = v*costheta*T
T = x/(v*cos(theta))......(1)
along vertical
______________
v0y = v*sin(theta)
acceleration ay = -g = -9.8 m/s^2
from equation of motion
y-y0 = vy*T + 0.5*ay*T^2
y-y0 = (v*sin(theta)*x)/(v*cos(theta)) - (0.5*g*x^2)/(v^2*(cos(theta))^2)
y-y0 = x*tan(theta) - ((0.5*g*x^2)/(v^2*(cos(theta))^2))
for x = 10 m
v = ?
theta = 44
y = H = 3.05 m
yo= h = 1.86 m
3.05 - 1.86 = 10*tan44 - (0.5*9.8*10^2/(v^2*(cos44)^2))
v = 10.6 m <<-----ANSWER
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