Gs Outlook.com-e X sa Mechanical Eng x m Blackboard Lear x yhttps://bbappsr X (-

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Gs Outlook.com-e X sa Mechanical Eng x m Blackboard Lear x yhttps://bbappsr X (--a c https://bbappsn.pmu.edu.sa/bbcswebdav/pid-297335-dt-content-rid-1154043-1/oourses/GEEN2311_103-1610/ G Two Forces ActPF How to Expresst New Tab 01.pdf Q.3 Express the force as a Cartesian vector F 750 N 45 EN .. :::45 l.. .11 10/2/2015 S51 AM Gs Outlook.com x G Mechanical E x BlockboardL. x L https://bbap: https://bbap xy ] w v ta w re x B Two Forces i xGChegg Study x New Tab d https://bbappsr pmu.edu.sa/bbswebdav/p d 297335-dt-content-rid-1154043 1/oourses/GEEN2311_103 1610/ %201.pdf Q.5 Determine the angle between the force and the line AB 4 m F=600 N 5:58 AM EN 10/2/2015

Explanation / Answer

Question 3:

Z - component of force will be = 750 * sin(45) = 530.33 N (Along positive z - axis)

XY planar component of the force will be 750 * cos(45) = 530.33 N

X - componet of the XY planar component = - 530.33 * cos(60) = - 265.165 N (Along the negative x - axis.)

Y - componet of the XY planar component = - 530.33 * sin(60) = - 459.28 N (Along the negative y - axis)

Hence, the cartesean vector will be -265.165 i - 459.28 j + 530.33 k

(Note, I assumed the the directions downwards and left as negatives, if otherwise the values of and x and y components will be positive)

Question 5:

Now, observe the triangle ABC,

length AB = sqrt(4^2 + 3^2) = 5 m

length AC = 5 m

length BC = sqrt(4^2 + 4^2) = 4 * sqrt(2)

using cosine law

cos(theta) = (AC^2 + AB^2 - BC^2) / (2 * AC * AB)

cos(theta) = (25 + 25 - 32) / (2 * 5 * 5) = 0.36

=> theta = 68.9 degrees.

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