17–42.The uniform crate has a mass of 50 kg and rests on the carthaving an inclined surface. Determine the smallestacceleration that will cause the crate either to tip or sliprelative to the cart. What is the magnitude of thisacceleration? The coefficient of static friction between thecrate and the cart is .ms= 0.5SOLUTIONEquations of Motion: Assume that the crate slips, then .a(1)(2)(3)Solving Eqs. (1), (2), and (3) yieldsAns. Since , then crate will not tip. Thus, the crate slips. Ans. x 6 0.3 ma = 2.01 m>s2N = 447.81 N x = 0.250 mR

a = – 2.01m/s²

The magnitude of the acceleration is 2.01m/s²

Explanation:

Given m = 50kg, θ = 15°, μs = 0.5

The forces acting on the crate are

The weight W = mg

The horizontal force F

The static frictional force

The reaction at the surface of the cart R

Taking x-axis to be parallel to the surface of the cart and y-axis to be perpendicular to the surface.

Resolving the forces into components parallel and perpendicular to the surface of the cart we have that

Wx = mgSinθ = 50×9.8×Sin15° = +126.8N

Wy = mgCosθ = 50×9.8Cos15° = –473.3N

Rx = 0

Ry = R,

(Ff)x = Ff = –μs×R = –0.5R

(Ff)y = 0

Fx = maCosθ = 50a×Cos15° = –48.3a

Fy = maSinθ = 50a×Sin15° = –12.94a

By Newtown's first law the sum of all the forces is zero

Summing all x-components forces

Rx +Wx + Ffx + Fx = 0

0 + 126.8 –0.5R – 48.3a = 0

0.5R +48.3a = 126.8 (1)

Summing all y-component forces

Ry +Wy + Ffy + Fy = 0

R – 473.3 + 0 –12.94a = 0

R = 12.94a + 473.3

Substituting R in equation in (1)

0.5(12.94a + 473.3) + 48.3a = 126.8

6.47a + 236.7 + 48.3a = 126.8

54.77a = 126.8 – 236.7

54.77a = – 109.9

a = – 109.9/54.77 = – 2.01m/s²

a = – 2.01m/s²

d. air pressure decreases as altitude increases.