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3 years ago

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A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to

the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?

1 answer:

rodikova [14]3 years ago

8 0

Answer:F=40.09 N

Explanation:

Given

weight of crate $W=50 N$

Inclination $\theta =37^{\circ}$

Frictional Force $f=10 N$

as the crate is moving with constant velocity therefore net Force on crate is zero

$F-50\sin (37)-f=0$

$F=50\sin (37)+10$

$F=30.09+10$

$F=40.09 N$

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vi = Initial speed = 0, ti = initial time = 0

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We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

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Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

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Normal force calculation:

Newton's first law

$sum Fy= 0$

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$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

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