A 0.53 kg arrow leaves a bowstring at a velocity of 63 m/s. If the arrow was initially at rest and then the string applied a for
1 answer:
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Answer:
Therefore, the force supported by the hydraulic cylinder is = -70.01 lb
Thus, the force supported by the bar = -233.84 lb
The reaction force supported by the reaction of the roller = 341.31 lb
Explanation:
The attached diagrams is shown in the file below:
From there; taking moments about point A
- 40 (180) - (80) Cos 37° + 55
-7200 + (-80 Cos 37° + 55 sin 37° ) = 0
= - 7200 - 30.79
- 30.79 = 7200
=
= -233.84 lb
Thus, the force supported by the bar = -233.84 lb
Taking the equilibrium of forces in the vertical direction
sin 37 +
cos 20 -
= 0
- 233.84 sin 37 + cos 20 - 180 = 0
cos 20 = 320.73
=
= 341.31 lb
The reaction force supported by the reaction of the roller = 341.31 lb
Taking the equilibrium of forces on the horizontal direction.
cos 37 -
sin 20° = 0
-233.84 cos 37 - + 341.31 sin 20° = 0
-186.75 - + 116.74 = 0
- -70.01 = 0
- = 70.01
= - 70.01 lb
Therefore, the force supported by the hydraulic cylinder is = -70.01 lb
Answer:
a) The object must have constant velocity.
d) The object must have zero acceleration.
Explanation:
We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:
where
F is the net force
m is the mass of the object
a is the acceleration
In this problem, the net force on the object is zero:
F = 0
This means that the acceleration of the object is also zero, according to the previous equation:
a = 0
So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:
Which means that , so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:
b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero
c)The object must be at the origin. --> false, since the object can be in motion