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2 years ago

A rope hangs between two cliffs, edge A and edge B. There is a 250 kg hiker sliding across and is closer to

Physics

1 answer:

iVinArrow [24]2 years ago

4 0

Answer:

1655.5 N

Explanation:

We resolve the forces on the rope into horizontal and vertical components. Since the only vertical force exerted on the rope is the weight of the man, this weight equals the vertical components of the tensions on both sides of the rope .

The only horizontal forces acting on the rope are horizontal components of both tensions on both sides. Since the hiker doesn't move sideways, they cancel out and he is balanced. So, the net horizontal force is zero.

Let T be the tension in the rope. So, each vertical component on side A is Tsin35° and that on side B is Tsin65°. Let the weight of the man be W = mg where m = mass of hiker = 250 kg and g = acceleration due to gravity = 9.8 m/s².

So, since the hiker balances, Tsin35° + Tsin65° = mg

0.5736T + 0.9063T = 250 kg × 9.8 m/s²

1.4799T = 2450 N

dividing both sides by 1.4799, we have

T = 2450 N/1.4799

T = 1655.52 N

T ≅ 1655.5 N

So, the tension in the rope equals T = 1655.5 N

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2 years ago

Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work

LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

$\frac{\Delta x}{\Delta t}$

Analysing the units, meters (displacement) and seconds (time) are basic units:

$\frac{m}{s}$

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

$F = ma$

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

$a=\frac{\Delta v}{\Delta t}$

Analysing the units:

$\frac{\frac{m}{s}}{s}=\frac{m}{s^2}$

Therefore, the unit of force is:

$kg\frac{m}{s^2}$

Pressure:

Pressure is given by the equation:

$P=\frac{F}{S}$ where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

$[l^2]=m^2$ , the unit of area

Dividing the aforementioned unit of force by the unit of area:

$\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}$ , the unit of pressure

Work:

Work is given by the equation:

$W=\vec{F}\cdot \vec{x}$ , (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

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2 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

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3 years ago