A 1170-kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure (Figure 1) . The ca
ble makes an angle of 31.0 ∘ above the surface of the ramp, and the ramp itself rises at 25.0 ∘ above the horizontal.
Draw a free-body diagram for the car.
Draw a free-body diagram for the car.
1 answer:
Refer to the diagram shown below.
The mass of the car is 1170 kg, therefore its weight is
W = (1170 kg)*(9.8 m/s²) = 11466 N
The component of the weight acting down the incline is
F = W sin(25°) = (11466 N)*sin(25°) = 4845.7 N
The normal reaction from the inclined plane is
N = W cos(25°) = (11466 N) cos(25°) = 1039.2 N
T = tension in the cable, acting at 31° above the surface of the ramp.
The Free Body Diagram on the right shows all the forces (friction is ignored)
and they FDB is sufficient for determining the value of T which establishes equilibrium.
The mass of the car is 1170 kg, therefore its weight is
W = (1170 kg)*(9.8 m/s²) = 11466 N
The component of the weight acting down the incline is
F = W sin(25°) = (11466 N)*sin(25°) = 4845.7 N
The normal reaction from the inclined plane is
N = W cos(25°) = (11466 N) cos(25°) = 1039.2 N
T = tension in the cable, acting at 31° above the surface of the ramp.
The Free Body Diagram on the right shows all the forces (friction is ignored)
and they FDB is sufficient for determining the value of T which establishes equilibrium.
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Answer:
(a)Therefore the highest altitude attained by the object is =576 ft .
(b)Therefore the object takes 6 sec to fall to the ground.
Explanation:
Initial velocity: Initial velocity is a velocity from which an object starts to move.
u is usually used for notation of initial notation.
Final velocity: Final velocity is a velocity of an object after certain second from starting.
The final velocity is denoted by v.
Acceleration: The difference of final velocity and initial velocity per unit time
The S.I unit of acceleration is m/s².
(a)
Given that u= 128 ft\sec and g = 32 ft/sec².
At highest point the velocity of the object is 0 i.e v=0
Since the displacement is opposite to the gravity.
Therefore acceleration( a)= -g = -32 ft/sec².
To find the time this to happen we use the following formula
Here v=0
⇒0=128+(-32) t
⇒32t=128
⇒t = 4 sec
To determine the height we use the following formula
⇒s= 256 ft
Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .
(b)
At the highest point the velocity of the object is 0.
so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft
To determine the time to fall we use the following formula
⇒t=6 sec
Therefore the object takes 6 sec to fall to the ground.