Three 5.8-kg masses are located at points in the xy plane, as shown. What is the

Question

Three 5.8-kg masses are located at points in the xy plane, as shown. What is the magnitude of the resultant force (caused by the other two masses) on the mass at x = 0.40 m, y = 0? e. 2.5 ´ 10-8 N Three 5.6-kg masses are located at points in the xy plane as shown in the figure. What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? a. 2.7 ´ 10-8 N
b. 2.1 ´ 10-8 N
c. 1.8 ´ 10-8 N
d. 2.4 ´ 10-8 N
e. 2.9 ´ 10-8 N Three 5.6-kg masses are located at points in the xy plane as shown in the figure. What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? a. 2.7 ´ 10-8 N
b. 2.1 ´ 10-8 N
c. 1.8 ´ 10-8 N
d. 2.4 ´ 10-8 N
e. 2.9 ´ 10-8 N a. 2.7 ´ 10-8 N
b. 2.1 ´ 10-8 N
c. 1.8 ´ 10-8 N
d. 2.4 ´ 10-8 N
e. 2.9 ´ 10-8 N e. 2.9 ´ 10-8 N

Explanation / Answer

let the masses at (0,0) , (0,0.3) and (0.4,0) be m1 , m2 and m3 respectively . m1 = m2 = m3 = 5.8 Kg we have to calculate force on m3 r13 = distance between m1 and m3 = 0.4 m r23 = distance between m2 and m3 = ( 0.42 + 0.32 )1/2 = 0.5 m force on m3 due to m1 =F1 =G m1 m3 /r213 = 1.4 *10-8 N along x-axis force on m3 due to m2 =F2 =G m2 m3 /r223 = 8.97 * 10-9 N along the direction making an angle e x-axis component of F2 along x- axis = Fx = F2 cos e component of F2 along y- axis = Fy = F2 sin e so, total force on m3 along x-axis = fx = F1 + F2 cos e and total force on m3 along y-axis = fy = F2 sin e   so, net force on m3 FR = (fx+fy)1/2 --------------------------------------------------(i) cos e = 4/5 sin e = 3/5 so; fx =  F1 + F2 cos e = 2.12*10-8 N and fy = F2 sin e = 5.382 *10-9 N from equation (i):- so; FR = ( fx + fy )1/2 = 2.2 * 10-8    N

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