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:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
