Amover in a moving truck is suing a rope to pull a 400-pound box up a ramp that has an incline of 25o. what is the force needed to hold the box in a stationary position to prevent the box from sliding down the ramp?

F = -52.8 N

Step-by-step explanation:

Given :

Inclination angle : 25°

Weight, W = 400 lb

The force N will push the box against the ramp.

Therefore, Force F will prevent the box from sliding down the ramp.

F = W sin 25°

  = 400 x (-0.132)

  = -52.8 N ( negative sign shows that the force applied in the opposite direction )

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b