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:
James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down
Explanation:
Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards
so here we can say
Upwards force = downwards Force + weight of snow
while if we find the other force which is acting downwards
then for that force we can say that net torque must be balanced
so here we have
so here we have
so here we can say that upward force by which we push up is always more than the downwards force