jwilson86
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Jake drags a sled of mass 25 kg across the snow. The sled is attached to a rope that pulls with a force of 50 N at an angle of 32 degrees. If there is no friction, what is the normal force acting on the sled? ​

Respuesta :

skyluke89

Answer:

218.5 N

Explanation:

In order for the sled to be in equilibrium along the vertical direction, the forces acting along this direction must be balanced. So the equilibrium equation is:

[tex]N+F sin \theta - mg =0[/tex]

where

N is the normal force

F = 50 N is the force that pulls the sled

[tex]\theta[/tex] is the angle between the force and the horizontal, so

[tex]F sin \theta[/tex] is the component of F acting along the vertical direction

(mg) is the weight of the sled, with

m = 25 kg being the mass of the sled

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the formula for N, we find

[tex]N=mg-F sin \theta = (25 kg)(9.8 m/s^2)-(50 N)sin 32^{\circ}=218.5 N[/tex]

temdan2001

By resolving into y - component, the normal force acting on the sled is 218.5 N.

FORCE

Force is the product of mass and acceleration. Where a frictional force is negligible, the resultant force will be equal to the force applied.

Given that Jake drags a sled of mass 25 kg across the snow. The sled is attached to a rope that pulls with a force of 50 N at an angle of 32 degrees.

If g is taken to be 9.8 m/s^2, the forces acting on the mass are:

The weight mg = 25 x 10 = 245 N

The normal reaction N = ?

The applied force F which can be resolved into x and y components

x - component = 50 x cos 32 = 42.4N

y - component = 50 x sin 32 = 26.5 N

If there is no friction, the normal force acting on the sled can be calculated by resolving all the forces into x and y component.

By resolving into y - component,

N + 26.5 = 245

N = 245 - 26.5

N = 218.5 N

Therefore, the normal force acting on the sled is 218.5 N.

Learn more about Force here: https://brainly.com/question/25239010

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