a) Force F(9.84 N) i - (2.95 N) k acts on a pebble with position vector r= (9.93

Question

a) Force F(9.84 N) i - (2.95 N) k acts on a pebble with position vector r= (9.93 m) j - (8.16m) k , relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (3.62 m, 0, -3.23 m)?

b) A 83 kg window cleaner uses a 12 kg ladder that is 4.6 m long. He places one end on the ground 2.1 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 1.4 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

c) In the figure, suppose the length L of the uniform bar is 2.7 m and its weight is 230 N. Also, let the block's weight W= 280 N and the angle ? = 39?. The wire can withstand a maximum tension of 350 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical
components of the force on the bar from the hinge at A?

Explanation / Answer

1)First Problem

Given that

F = ( 9.84i + 0 j - 2.95 k ) N

Position vector relative to origin

r1 = ( 0i + 9.93j - 8.16k) m

Position vector relative to coordinates (3.62 m, 0, -3.23 m)

r2 = ( 0i + 9.93j - 8.16k) m - ( 3.62i + 0j - 3.23k) m

= ( - 3.62i + 9.93j - 4.93k) m

(a) Torque = F x r1

= ( 9.84i + 0 j - 2.95 k ) N x ( 0i + 9.93j - 8.16k) m

= (-24.072i +80.29j + 97.71) N-m

(b) T = F x r2

= ( 9.84i + 0 j - 2.95 k ) N x ( - 3.62i + 9.93j - 4.93k) m

= (29.29 + 59.19j + 97.71k ) N-m

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conserving moment from the bottom of ladder
late F be the force acted on window
hence
angel made by ladder with horizontal is =cosinverce 2.1/4.6 =62.84 degree
F*4.6 sin62.84 =12 *9.8*1 +83*9.8*1.4cos62.84

hence F =123.29 N

Y force on bottom is (83+12)*9.8 =931 N

X force =123.29 N

magnitude = 939.12 N

angle =taninverce(931/123.29) =82.46 degrees

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3)

a)

in equilibrium net torque = 0

T*sintheta*L = W1*L/2 + W*x

350*sin39*2.7 = 230*(2.7/2) + 280*x

x = 1.02 m ,<---------------answer

++++++++++++

b)

along horizantal

Fnet = 0

Fx = T*cos33

Fx = 350*cos39 = 272 N <<--------answer

c)

along vertical

Fnet = 0

Fy + T*sin33 - W1 - W = 0

Fy = -350*sin39 + 230 + 280

Fy = 289.74 N <<<--------answer

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