Two forces, 354 N at 9 ? and 311 N at 25 ? are applied to a car in an effort to

Question

Two forces, 354 N at 9? and 311 N at 25? are applied to a car in an effort to accelerate it.

3294 kg

9? 25?

A) What is the magnitude of the resultant of these two forces?
Answer in units of N.

B)Find the direction of the resultant force (in relation to forward, with counterclockwise considered positive).

Answer in degrees from the positive x-axis, with counter-clockwise positive, within the limits of ?180? to 180? .
Answer in units of ?.

C) If the car has a mass of 3294 kg, what acceleration does it have?

Ignore friction.

354 N

311 N

Explanation / Answer

To get the resultant of these two forces convert the polar notation to rectangular notation and add the x values and then the y values. Convert this back to polar notation.

354N @ 9 degrees = 354N*(Cosine 9) +j 354N*(sine 9) = 349.64 N+ j 55.38 N

311N @ 25degrees = 311N*(Cosine 25) +j 311N*(sine 25) = 281.86N +j131.4N

Add the x+jy terms to get = 631.5+ j*186.78
x= 631.5 N

y= 186.78 N
Convert to polar.

Resultant in polar = Square root of (631.5^2 + 186.78^2 m) = 658.54 N


direction given by tan(theta) = 186.78/631.5=> theta = 16.48 deg

if total force = 354 N, then resultant accel = F/m = 354/3294 kg = 0.107m/s/s

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