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spayn [35]
3 years ago
5

What is the force required to produce a 1.4 Nm torque when applied to a door at a 60.0º angle and 0.40 m from the hinge?

Physics
1 answer:
Marysya12 [62]3 years ago
5 0
Refer to the diagram shown below.

The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N

Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m

This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N

Answer: 4.04 N

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F= 4788 N

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Because the car moves with uniformly accelerated movement we apply the following formula:

vf²=v₀²+2*a*d Formula (1)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s    

vf: final speed in m/s  

a: acceleration in m/s²

Data

d=36.9 m

v₀=14.0 m/s m/s    

vf= 0  

Calculating of the acceleration of the car

We replace dta in the formula (1)

vf²=v₀²+2*a*d

(0)²=(14)²+2*a*(36.9)

-(14)²= (73.8) *a

a= - (196) /  (73.8)

a= - 2.66 m/s²

Newton's second law of the car in direction  horizontal (x):

∑Fx = m*ax Formula (2)

∑F : algebraic sum of the forces in direction x-axis (N)

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a : acceleration  (m/s²)

Data

m=1800 Fkg

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Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :

We replace data in the formula (2)

-F= (1800 kg) * ( -2.66 m/s² )

F= 4788 N

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