To move a large crate across a rough floor, you push on it with a force F at an angle of 21....F=250 N)

Question

To move a large crate across a rough floor, you push on it with a

force F at an angle of 21° below the horizontal, as shown in FIGURE

6-34. Find the force necessary to start the crate moving, given that

the mass of the crate is 32 kg and the coefficient of static friction

between the crate and the floor is 0.57. (F=250N)

Answer 1

I'm assuming the figure looks like this:

Draw a free body diagram:

F_F is the force of friction. W is the weight of the crate (mass_crate•g). N is the Normal Force. . Fx is the component of the Force F in question, in the x-direction and Fy is the component of the Force F in question, in the y-direction.

W = (32)(9.8) = 313.6 N

Fx = Fcos(21°)

Fy = Fsin(21°)

F_F = μN = 0.57N

ΣF_y = N - W - F_y = N - 313.6 - Fsin(21°)

ΣF_x = F_x - F_F = Fcos(21°) - 0.57N

Because there is no motion in the y-direction, ΣF_y = 0 meaning N - 313.6 - Fsin(21°) = 0 or:

N - 313.6 = Fsin(21°) [Equation A]

Just at the point when the crate "breaks free of the floor", there is no motion in the x-direction meaning ΣF_x = 0 meaning Fcos(21°) - 0.57N = 0 or:

0.57N = Fcos(21°) [Equation B]

To solve, divide each side of Equation A by the corresponding side of Equation B:

(N - 313.6)/0.57N = Fsin(21°)/Fcos(21°)

N/0.57N - 313.6/0.57N = sin(21°)/cos(21°)

1/0.57 - 313.6/0.57N = tan(21°)

1.7544 - 313.6/0.57N = 0.3839

- 313.6/0.57N = - 1.3705

313.6 = 1.3705(0.57N)

313.6 = 0.7812N

N = 401.43 Newtons

Using Equation B, 0.57N = Fcos(21°) we can say F = 0.57N/cos(21°) = 0.57(401.43)/cos(21°) = 245.097 N

Since there are only 2 significant figures in the "givens", we round our answer to 2 sig figs so F = 250 N