Respuesta :
Answer:
Explanation:
The first expression ; five over six is subtracted from two whole number two over 5 and the results is multiplied by three quarter
From the diagram where Fw = the force of the gravity, Fn = the normal force, Ff = the force of friction.
since it is said that ; the block is not moving, the sum of the y components of the forces acting on the block must be zero. hence sum of y components ( summation Fy) = 0 ; Ffsinθ + Fncosθ - Fw = 0
The attached file shows the diagram
Answer:
A) F_n*cos(Q) + F_f*sin(Q) - F_w = 0
B) - F_n*sin(Q) + F_f*cos(Q) = 0
C) F_n = F_w*cos(Q)
Explanation:
Part A:
Find an expression for the sum of the y components of the forces acting on the block Express your answer in terms of some or all of the variables F_n, F_f, and F_w and Q.
Solution:
- The sum of forces in y direction in coordinate system b is:
F_n*cos(Q) + F_f*sin(Q) - F_w = 0
Part B:
Find an expression for the sum of the x components of the forces acting on the block Express your answer in terms of some or all of the variables F_n, F_f, and Q.
Solution:
- The sum of forces in x direction in coordinate system b is:
- F_n*sin(Q) + F_f*cos(Q) = 0
Part C:
Using the equations you found in the two previous parts, find an expression for F_n using F_w and Q not F_f:
Solution:
- Using part B, The expression for F_f is:
F_f = F_n*tan(Q)
- Substitute the above Eq in part A:
F_n*cos(Q) + F_n*tan(Q)*sin(Q) - F_w = 0
F_n*(cos(Q) + tan(Q)*sin(Q)) = F_w
F_n*(sec(Q)) = F_w
F_n = F_w*cos(Q)
