Can anyone explain this problem with an aid of a diagram?

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2 Answers
Jan 27, 2017

Answer:

The resultant force is 86.74 N at an angle of 72.1°

Sorry, no diagram!

Explanation:

First, you will resolve each vector (given here in standard form) into rectangular components (#x# and #y#).

Then, you will add together the #x-#components and add together the #y-#components. This will given you the answer you seek, but in rectangular form.

Finally, convert the resultant into standard form.

Here's how:

Resolve into rectangular components (note the changes of angle measure into standard angles):

#Fx_1 = 35 cos 110° = 35 (-0.342) = -11.97 N#

#Fy_1 = 35 sin 110° = 35 (0.940) = 32.89 N#

#Fx_2 = 60 cos 90° = 0 N#

#Fx_1 = 60 sin 90° = 60 N#

#Fx_3 = 40 cos -15° = 40 (0.969) = 38.64 N#

#Fy_3 = 40 sin -15° = 40 (-0.259) = -10.35 N#

Now, add the one-dimensional components

#F_x = F_(x1)+F_(x2) +F_(x3) = 26.67 N#

and

#F_y = F_(y1)+F_(y2) +F_(y3) = 82.54 N#

This is the resultant force in rectangular form. With a positive #x#-component and a positive #y#-component, this vector points into the 1st quadrant.

Now, convert to standard form:

#F = sqrt((F_x)^2+(F_y)^2) = sqrt((26.67)^2+82.54^2) = 86.74 N#

#theta=tan^(-1)(82.54/(26.67)) = 72.1°#

Jan 27, 2017

Answer:

#"Problem was solved again."#

Explanation:

#"all forces acting on object"#

enter image source here

#".................................................................................................."#

#"The Force "F_1" and its components(vertical,horizontal)"#

enter image source here

#F_("1x")=-F_1.sin(20)=-35.0,34202014=-11.97" N"#

#F_("1y")=F_1.cos(20)=35.0,93969262=32.89" N"#

#"...................................................................................."#

#"The Force "F_2" and its components(vertical,horizontal)"#

#"The Force "F_2 " has vertical component only ."#

#F_("2x")=0#

#F_("2y")=60" N"#

enter image source here

#"...................................................................................................."#

#"The Force "F_3" and its components(vertical,horizontal)"#

enter image source here

#F_("3x")=F_3.sin(75)=40.0,96592583=38.64" N "#

#F_("3y")=-F_3.cos(75)=-40.0,25881905=-10.35" N"#

#.....................................................................................................#

#"now let us find the total "F_x" and "F_ y " components."#

#F_x=F_("1x")+F_("2x")+F_("3x")#

#F_y=F_("1y")+F_("2y")+F_("3y")#

#F_x=-11.97+0+38.64=26.67" N"#

#F_y=32.89+60-10.35=82.54" N"#

#".................................................................................."#

#"Resultant Vector..."#

enter image source here

#" magnitude of the resultant vector can be calculated :"#

#F=sqrt((F_x)^2+(F_y)^2)#

#F=sqrt((26.67)^2+(82.54)^2)#

#F=sqrt(711.2889+6812.8516)#

#F=sqrt(7524.1405)#

#F=86.74" N"#

#"we must find " tan(theta) " for direction of the resultant vector."#

#tan(theta)=(F_x)/(F_y)#

#tan(theta)=(26.67)/(82.54)#

#tan(theta)=0.32311606#

#theta=17.91 #