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Maru [420]
3 years ago
14

A jumbo jet weighing 65,000 kg Is accelerating at 3 m/s^2 while its jet engines are putting out 200000 N of thrust.

Physics
1 answer:
sveticcg [70]3 years ago
8 0

Answer:

Fa = 5000 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

Let's assume that the movement of the plane is to the right, any movement or force to the right will be marked with a positive sign, while any force or movement to the left, will be taken as negative.

The force of the turbine drives the plane to the right, therefore it is positive, the acceleration is constant and keeps the movement to the right, therefore it is positive, the wind drag force tries to prevent the movement of the plane to the left therefore it is negative, with this analysis we deduce the following equation.

ΣF = m*a

where:

ΣF = sum of forces [N] (units of Newtons)

m = mass = 65000 [kg]

a = acceleration = 3 [m/s²]

Fa = force exerted by the air [N]

200000 - Fa = 65000*3

Fa = 200000 - (3*65000)

Fa = 5000 [N]

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A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable
Nuetrik [128]

Answer:

a) 29062.125 N·m

b) 0 N·m

c) Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

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Answer:

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