2. Two people are liftinga heavy box (625 N). One person is applying a force of

Question

2. Two people are liftinga heavy box (625 N). One person is applying a force of 400 N at anangle
of 40 degrees from the horizontal. The other person is applying aforce of 320 N. At what angle
should the second force be applied so the box will acceleratestraight up? How fast will the
object be moving after 1 second?

If the box is being pulledstraight up, it is in equilibrium in the x-direction and
accelerating in the y-direction.

x-direction:
Sum Fx = 0
400 cos(40) + 320 cos(angle) = 625 cos(270) = 0 CAN YOUEXPLAIN?
Angle = 163.2 degrees.

y-direction:
Sum Fy = m*ay
W = mg … m = 625 / 9.8 = 63.8 kg
400 sin(40) + 320 sin(163.2) + 625 sin(270) = (63.8) ay
-275.39/63.8 = -4.3
ay = - 4.3 m/s²

The negative signindicates that the box is not being lifted, as it is acceleratingin a
downward direction.

Using kinematics:
ay = (Vfy = Viy) / t
-4.3 = (Vfy – 0) / 1
Vfy = -4.3 m/s
The box has a velocity of 4.3 m/s downward after 1second.

Explanation / Answer

If the box is beingpulled straight up, it is in equilibrium in the x-direction and
accelerating in the y-direction.

x-direction:
Sum Fx = 0
400 cos(40) + 320 cos() =0           (let the force 320 be at an angle with the+ve x axis also only these 2 forces act on it on x axis)
= 163.2 degrees.

y-direction:
Sum Fy = m*ay
W = mg … m = 625 / 9.8 = 63.8 kg
400 sin(40) + 320 sin(163.2)- 625 = (63.8) ay
-275.39/63.8 = -4.3
ay = - 4.3 m/s²

The negative signindicates that the box is not being lifted, as it is acceleratingin a
downward direction.

Using kinematics:
ay = (Vfy - Viy) / t
-4.3 = (Vfy – 0) / 1
Vfy = -4.3 m/s
The box has a velocity of 4.3 m/s downward after 1second.

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