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A box at rest on a ramp is in equilibrium, as shown.

What is the force of static friction acting on the box? Round your answer to the nearest whole number.
_______N

What is the normal force acting on the box? Round your answer to the nearest whole number.
_______ N

A box at rest on a ramp is in equilibrium as shown What is the force of static friction acting on the box Round your answer to the nearest whole number N What i class=

Respuesta :

MathPhys

Answer:

Ffs = 251 N

Fn = 691 N

Explanation:

Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.

The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:

Fgx = Fg sin 20°

Fgy = Fg cos 20°

Sum of the forces in the x direction (parallel to the ramp):

∑F = ma

Ffs − Fgx = 0

Ffs = Fgx

Ffs = Fg sin 20°

Ffs = 735 sin 20°

Ffs ≈ 251

Sum of the forces in the y direction (normal to the ramp):

∑F = ma

Fn − Fgy = 0

Fn = Fgy

Fn = Fg cos 20°

Fn = 735 cos 20°

Fn ≈ 691

To answer that question we need to apply equations of movement ( from Newton´s laws )

In equilibrium:

∑F  = 0         or    ∑Fx = 0     ;  ∑Fy = 0

Solution is:

a) F(sf) = 251 [N]

b) Fn = 691 [N]

From the attached drawings we can see:  ( Body free diagram)

∑ Fₓ  = F(sf)  - Pₓ  = 0           where P = m×g  = 735 [N] ( the weigth)

and Pₓ = P× cos20°

Then     F(sf)  = Pₓ × sin 20°

F(sf) = m×g×cos20°  =  735× 0.34202 [N]

F(sf) = 251.3847

F(sf) = 251 [N]

rounding to the nearest number

F(sf) = 691 [N]

∑ Fy = 0

∑ Fy = Fn - Py  = 0                     Py = P×cos20°      Py = m×g×cos20°

Py = 735×0.939693 [N]     Py =   [N]

Fn = Py = 690.674 [N]

rounding to the nearest whole number

Fn = 691 [N]

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