Forces And Friction

Question 1

Find the component of each force in the x-direction

( Diagram shows the x-axis and y-axis directions )
( The 100 N force is in the 3rd quadrat pointing downwards )
( The arrow is 18 degrees from the negative y-axis line )

Answer 1

• ( To work out the force in the x-direction, we find out the value of the magnitude on the x-axis )
• ( Using trigonometry )
• 90 - 18 = 72
• ( The arrow is 72 degrees from the negative x-axis direction )
• ( We form a triangle against the arrow, the 100 N becomes the hypotenuse and the x-axis is the adjacent )
• • A = 100 x cos( 72 )
• • A = 30.90169944
• A = - 30.90 N
• ( The adjacent was initially negative because the magnitude was in the negative direction of x )

Question 2

Find the component of each force in the y-direction

( Diagram shows the x-axis and y-axis directions )
( The 100 N force is in the 3rd quadrat pointing downwards )
( The arrow is 18 degrees from the negative y-axis line )

Answer 2

• ( The 100 N arrow is 18 degrees from the negative y-axis )
• ( Which makes the y-axis the adjacent and the 100N arrow the hypotenuse )
• • A = 100 x cos( 18 )
• • A = 95.10565163
• A = - 95.10 N

Question 3

Hence write each force in the form pi + qj where i and j are the unit vectors in the x and y directions respectively.

( Magnitude in x-direction = - 30.90 )
( Magnitude in y-direction = - 95.10 )

Answer 3

• ( - 30.90 i - 95.10 j ) N

Question 4

Find the magnitude and direction of the resultant force acting on each of the particles shown below.

( The diagram shows the x-axis )
( There are two forces acting on the particle, away from the x-axis )
( The arrow going towards the left has a force of 25 N and is 50 degrees from the x-axis )
( The arrow going towards the right has a force of 35 N and is 30 degrees from the x-axis )
( We labeled the angle from R N to the x-axis, k )

Answer 4

• ( In this question, you must show a resultant force on the particle from the two forces acting on it )
• ( The resultant force will be closer the the arrow holding a greater degrees from the axis )
• ( Next you move either the 35 N or 25 N arrow to the corresponding arrow, passing the resultant force arrow, making sure that it’s parallel to the original line )
• ( In this case, I moved the 35 N arrow to the 25 N arrow )
• ( You will be able to identify that a “ C “ shape has been made with the arrows )
• ( We know that angles within the “ C “ shape add up to 180 degrees )
• ( One of the angles within the “ C “ is 100 degrees, so the other is 80 degrees )
• ( Using the forces 25 N, the new 35 N and R N, we form a right angles triangle, which contains the 80 degrees )
• ( Label the angle on R N, theta )
• ( We use the cosine rule to find R )
• R^2 = ( 25 )^2 + ( 35 )^2 - 2( 25 ) ( 35 ) cos ( 80 )
• R^2 = 1546.115689
• R = 39.32067763
• ( To find out the degrees of R N from the x-axis ( k ), we need to find theta )
• ( We use the sine rule to find theta )
• sin( theta ) / 35 = sin( 80 ) / 39.32…
• Theta = sin^-1 ( 35 x sin( 80 ) / 39.32… )
• Theta = 61.23419385
• 180 - ( 61.2… + 50 ) = k
• k = 68.76580615
• k = 68.77 degrees
• Magnitude = 39.32 N
• Direction of resultant force = 68.77 degrees